Questions and Answers (Board Exam)

Introduction to Board Exam Questions

This section contains 30 highly important board-style questions based on standard Class 12 three-dimensional geometry patterns. In board exams, presentation is as important as the final answer. Always state the formula first, identify clearly whether you are working with a point, a position vector, a direction vector, a normal vector, or direction ratios, and then substitute values carefully. In 3D geometry, many errors happen because students mix up the coordinates of a point with the components of a direction vector, so write each step neatly.

Question 1 [CBSE 2026]

Find the direction cosines of the line $\frac{x-1}{2} = \frac{y+2}{-3} = \frac{z-4}{6}$.

Solution: Step 1: Compare the given line with the standard symmetric form $$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}.$$ Hence, the direction ratios of the line are $$a=2, \quad b=-3, \quad c=6.$$

Step 2: Direction cosines are obtained by dividing each direction ratio by the magnitude of the direction vector. So first compute $$\sqrt{a^2+b^2+c^2} = \sqrt{2^2+(-3)^2+6^2} = \sqrt{4+9+36} = \sqrt{49} = 7.$$

Step 3: Therefore, $$l = \frac{2}{7}, \quad m = -\frac{3}{7}, \quad n = \frac{6}{7}.$$

Step 4: Check the identity for direction cosines: $$\left(\frac{2}{7}\right)^2 + \left(-\frac{3}{7}\right)^2 + \left(\frac{6}{7}\right)^2 = \frac{4+9+36}{49} = 1.$$ So the answer is correct.

Answer: $\frac{2}{7}, -\frac{3}{7}, \frac{6}{7}$

Question 2 [CBSE 2025]

Find the Cartesian equation of the line passing through the points $(3, -2, -5)$ and $(3, -2, 6)$.

Solution: Step 1: Let the given points be $$A(3,-2,-5), \quad B(3,-2,6).$$ The line through two points has direction ratios $$a = x_2-x_1, \quad b = y_2-y_1, \quad c = z_2-z_1.$$ So, $$a = 3-3=0, \quad b = -2-(-2)=0, \quad c = 6-(-5)=11.$$

Step 2: Use the two-point symmetric form $$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}.$$ Substituting $A(3,-2,-5)$ and direction ratios $(0,0,11)$, $$\frac{x-3}{0} = \frac{y+2}{0} = \frac{z+5}{11}.$$

Step 3: Interpret the zero denominators correctly. This means $$x-3=0 \implies x=3,$$ $$y+2=0 \implies y=-2.$$ So the line is parallel to the z-axis and passes through $(3,-2,-5)$.

Answer: $\frac{x - 3}{0} = \frac{y + 2}{0} = \frac{z + 5}{11}$

Question 3 [CBSE 2024]

Find the shortest distance between the lines $\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k})$.

Solution: Step 1: Compare with the standard vector forms $$L_1: \vec{r}=\vec{a}_1+\lambda\vec{b}_1, \qquad L_2: \vec{r}=\vec{a}_2+\mu\vec{b}_2.$$ Thus, $$\vec{a}_1 = \hat{i}+2\hat{j}+\hat{k}, \quad \vec{b}_1 = \hat{i}-\hat{j}+\hat{k},$$ $$\vec{a}_2 = 2\hat{i}-\hat{j}-\hat{k}, \quad \vec{b}_2 = 2\hat{i}+\hat{j}+2\hat{k}.$$

Step 2: Find the vector joining the known points on the two lines: $$\vec{a}_2-\vec{a}_1 = (2-1)\hat{i}+(-1-2)\hat{j}+(-1-1)\hat{k} = \hat{i}-3\hat{j}-2\hat{k}.$$

Step 3: Use the shortest distance formula for skew lines: $$d = \left|\frac{(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)}{|\vec{b}_1\times\vec{b}_2|}\right|.$$ So first compute the cross product: $$\vec{b}_1\times\vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix}.$$ Expanding, $$\vec{b}_1\times\vec{b}_2 = \hat{i}((-1)(2)-1(1)) - \hat{j}(1\cdot2-1\cdot2) + \hat{k}(1\cdot1-(-1)\cdot2)$$ $$= -3\hat{i} + 0\hat{j} + 3\hat{k}.$$

Step 4: Its magnitude is $$|\vec{b}_1\times\vec{b}_2| = \sqrt{(-3)^2+0^2+3^2} = \sqrt{18} = 3\sqrt{2}.$$

Step 5: Compute the scalar triple product in the numerator: $$(\hat{i}-3\hat{j}-2\hat{k})\cdot(-3\hat{i}+3\hat{k}) = -3 + 0 - 6 = -9.$$

Step 6: Therefore, $$d = \left|\frac{-9}{3\sqrt{2}}\right| = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}.$$

Answer: $\frac{3\sqrt{2}}{2}$ units

Question 4 [CBSE 2023]

Find the coordinates of the foot of the perpendicular drawn from the point $P(0, 2, 3)$ to the line $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$.

Solution: Step 1: Let the foot of the perpendicular be $Q$. A general point on the given line can be written by taking the common parameter as $\lambda$: $$x = 5\lambda - 3, \quad y = 2\lambda + 1, \quad z = 3\lambda - 4.$$ So, $$Q(5\lambda-3,\ 2\lambda+1,\ 3\lambda-4).$$

Step 2: The direction ratios of the given line are $$(5,2,3).$$ Since $PQ$ is perpendicular to the line, the vector $\overrightarrow{PQ}$ must be orthogonal to $(5,2,3)$. Now, $$\overrightarrow{PQ} = (5\lambda-3-0,\ 2\lambda+1-2,\ 3\lambda-4-3)$$ $$= (5\lambda-3,\ 2\lambda-1,\ 3\lambda-7).$$

Step 3: Apply perpendicularity condition: $$(5\lambda-3)(5) + (2\lambda-1)(2) + (3\lambda-7)(3) = 0.$$

Step 4: Simplify: $$25\lambda - 15 + 4\lambda - 2 + 9\lambda - 21 = 0$$ $$38\lambda - 38 = 0$$ $$\lambda = 1.$$

Step 5: Substitute back in the coordinates of $Q$: $$Q = (5(1)-3,\ 2(1)+1,\ 3(1)-4) = (2,3,-1).$$

Answer: $(2, 3, -1)$

Question 5 [CBSE 2026]

Find the value of $\lambda$ so that the lines $\frac{1-x}{3} = \frac{7y-14}{2\lambda} = \frac{z-3}{2}$ and $\frac{7-7x}{3\lambda} = \frac{y-5}{1} = \frac{6-z}{5}$ are at right angles.

Solution: Step 1: Rewrite both lines in standard symmetric form. For the first line, $$\frac{1-x}{3} = \frac{7y-14}{2\lambda} = \frac{z-3}{2}$$ becomes $$\frac{-(x-1)}{3} = \frac{7(y-2)}{2\lambda} = \frac{z-3}{2}.$$ Hence its direction ratios are $$a_1=-3, \quad b_1=\frac{2\lambda}{7}, \quad c_1=2.$$

For the second line, $$\frac{7-7x}{3\lambda} = \frac{y-5}{1} = \frac{6-z}{5}$$ becomes $$\frac{-7(x-1)}{3\lambda} = \frac{y-5}{1} = \frac{-(z-6)}{5}.$$ Thus its direction ratios are $$a_2=-\frac{3\lambda}{7}, \quad b_2=1, \quad c_2=-5.$$

Step 2: For two lines to be perpendicular, the dot product of their direction ratios must be zero: $$a_1a_2 + b_1b_2 + c_1c_2 = 0.$$ So, $$(-3)\left(-\frac{3\lambda}{7}\right) + \left(\frac{2\lambda}{7}\right)(1) + 2(-5) = 0.$$

Step 3: Simplify: $$\frac{9\lambda}{7} + \frac{2\lambda}{7} - 10 = 0$$ $$\frac{11\lambda}{7} = 10$$ $$11\lambda = 70$$ $$\lambda = \frac{70}{11}.$$

Answer: $\frac{70}{11}$

Question 6 [CBSE 2025]

Find the equation of the plane passing through the points $A(1, 1, -1), B(6, 4, -5),$ and $C(-4, -2, -3)$.

Solution: Step 1: Find two vectors lying in the plane: $$\vec{AB} = (6-1,\ 4-1,\ -5-(-1)) = (5,3,-4),$$ $$\vec{AC} = (-4-1,\ -2-1,\ -3-(-1)) = (-5,-3,-2).$$

Step 2: A normal vector to the plane is $$\vec{n} = \vec{AB} \times \vec{AC}.$$ So, $$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 3 & -4 \\ -5 & -3 & -2 \end{vmatrix}.$$ Expanding, $$\vec{n} = \hat{i}(3(-2)-(-4)(-3)) - \hat{j}(5(-2)-(-4)(-5)) + \hat{k}(5(-3)-3(-5))$$ $$= \hat{i}(-6-12) - \hat{j}(-10-20) + \hat{k}(-15+15)$$ $$= -18\hat{i} + 30\hat{j} + 0\hat{k}.$$ A simplified normal vector is $$(3,-5,0).$$

Step 3: Use the point-normal form through $A(1,1,-1)$: $$3(x-1) - 5(y-1) + 0(z+1) = 0.$$

Step 4: Simplify: $$3x - 3 - 5y + 5 = 0$$ $$3x - 5y + 2 = 0.$$

Step 5: Verify with the third point $C(-4,-2,-3)$: $$3(-4) - 5(-2) + 2 = -12 + 10 + 2 = 0.$$ So the equation is correct.

Answer: $3x - 5y + 2 = 0$

Question 7 [CBSE 2024]

Find the equation of the plane passing through the line of intersection of the planes $\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 6$ and $\vec{r} \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) = -5$, and passing through the point $(1, 1, 1)$.

Solution: Step 1: Convert both planes into Cartesian form: $$x+y+z=6 \implies x+y+z-6=0,$$ $$2x+3y+4z=-5 \implies 2x+3y+4z+5=0.$$

Step 2: The family of planes through their line of intersection is $$(x+y+z-6) + \lambda(2x+3y+4z+5)=0.$$

Step 3: Since the required plane passes through $(1,1,1)$, substitute this point: $$(1+1+1-6) + \lambda(2+3+4+5)=0$$ $$-3 + 14\lambda = 0$$ $$\lambda = \frac{3}{14}.$$

Step 4: Substitute this back: $$(x+y+z-6) + \frac{3}{14}(2x+3y+4z+5)=0.$$ Multiply throughout by 14: $$14(x+y+z-6) + 3(2x+3y+4z+5)=0.$$

Step 5: Simplify: $$14x+14y+14z-84 + 6x+9y+12z+15 = 0$$ $$20x + 23y + 26z - 69 = 0.$$

Answer: $20x + 23y + 26z = 69$

Question 8 [CBSE 2022]

Find the distance of the point $(2, 5, -3)$ from the plane $\vec{r} \cdot (6\hat{i} - 3\hat{j} + 2\hat{k}) = 4$.

Solution: Step 1: Convert the plane into Cartesian form: $$6x - 3y + 2z = 4 \implies 6x - 3y + 2z - 4 = 0.$$ Thus, $$A=6, \quad B=-3, \quad C=2, \quad D=-4.$$

Step 2: Use the point-to-plane distance formula: $$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2+B^2+C^2}}.$$ Substitute $(x_1,y_1,z_1)=(2,5,-3)$: $$d = \frac{|6(2) - 3(5) + 2(-3) - 4|}{\sqrt{6^2+(-3)^2+2^2}}.$$

Step 3: Simplify numerator and denominator: $$|12 - 15 - 6 - 4| = |-13| = 13,$$ $$\sqrt{36+9+4} = \sqrt{49} = 7.$$

Step 4: Therefore, $$d = \frac{13}{7}.$$

Answer: $\frac{13}{7}$ units

Question 9 [CBSE 2026]

Find the angle between the line $\frac{x+1}{2} = \frac{y}{3} = \frac{z-3}{6}$ and the plane $10x + 2y - 11z = 3$.

Solution: Step 1: The direction vector of the line is taken from the denominators: $$\vec{b} = (2,3,6).$$ The normal vector of the plane is taken from the coefficients of $x,y,z$: $$\vec{n} = (10,2,-11).$$

Step 2: If $\phi$ is the angle between the line and the plane, then $$\sin\phi = \frac{|\vec{b}\cdot\vec{n}|}{|\vec{b}|\,|\vec{n}|}.$$

Step 3: Compute the dot product: $$\vec{b}\cdot\vec{n} = 2(10) + 3(2) + 6(-11) = 20 + 6 - 66 = -40.$$ So, $$|\vec{b}\cdot\vec{n}| = 40.$$

Step 4: Compute magnitudes: $$|\vec{b}| = \sqrt{2^2+3^2+6^2} = \sqrt{49} = 7,$$ $$|\vec{n}| = \sqrt{10^2+2^2+(-11)^2} = \sqrt{225} = 15.$$

Step 5: Substitute: $$\sin\phi = \frac{40}{7\cdot 15} = \frac{40}{105} = \frac{8}{21}.$$ Therefore, $$\phi = \sin^{-1}\left(\frac{8}{21}\right).$$

Answer: $\sin^{-1}(8/21)$

Question 10 [CBSE 2023]

Show that the lines $\frac{x+3}{-3} = \frac{y-1}{1} = \frac{z-5}{5}$ and $\frac{x+1}{-1} = \frac{y-2}{2} = \frac{z-5}{5}$ are coplanar.

Solution: Step 1: Identify a point and direction ratios from each line. For the first line: $$(x_1,y_1,z_1)=(-3,1,5), \quad (a_1,b_1,c_1)=(-3,1,5).$$ For the second line: $$(x_2,y_2,z_2)=(-1,2,5), \quad (a_2,b_2,c_2)=(-1,2,5).$$

Step 2: The condition for coplanarity is $$\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0.$$ So, $$\begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix}.$$

Step 3: Expand the determinant: $$2(1\cdot 5 - 5\cdot 2) - 1((-3)\cdot 5 - 5(-1)) + 0$$ $$= 2(5-10) - (-15+5)$$ $$= 2(-5) - (-10) = -10 + 10 = 0.$$

Step 4: Since the determinant is zero, the two lines are coplanar.

Answer: Proved.

Question 11 [CBSE 2025]

Find the equation of the plane passing through the point $(1, -1, 2)$ and parallel to the plane $3x - 2y + 5z = 10$.

Solution: Step 1: A plane parallel to $$3x - 2y + 5z = 10$$ must have the same normal vector. So its equation is of the form $$3x - 2y + 5z = d.$$

Step 2: Since it passes through $(1,-1,2)$, substitute this point: $$3(1) - 2(-1) + 5(2) = d.$$

Step 3: Simplify: $$3 + 2 + 10 = d \implies d = 15.$$

Step 4: Hence the required plane is $$3x - 2y + 5z = 15.$$

Answer: $3x - 2y + 5z = 15$

Question 12 [CBSE 2024]

Find the vector equation of the line passing through the point $(1, 2, 3)$ and parallel to the line $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + t(2\hat{i} + 3\hat{j} - \hat{k})$.

Solution: Step 1: The required line passes through the point $(1,2,3)$, whose position vector is $$\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}.$$

Step 2: Since the required line is parallel to the given line, it has the same direction vector: $$\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}.$$

Step 3: Use the vector equation of a line through point $\vec{a}$ and parallel to vector $\vec{b}$: $$\vec{r} = \vec{a} + \lambda\vec{b}.$$ So, $$\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k}).$$

Answer: $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$

Question 13 [CBSE 2026]

Find the distance between the parallel lines $\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$ and $\vec{r} = (3\hat{i} + 3\hat{j} - 5\hat{k}) + \mu(2\hat{i} + 3\hat{j} + 6\hat{k})$.

Solution: Step 1: Identify the known position vectors and common direction vector: $$\vec{a}_1 = (1,2,-4), \quad \vec{a}_2 = (3,3,-5), \quad \vec{b}=(2,3,6).$$

Step 2: For parallel lines, use the formula $$d = \frac{|(\vec{a}_2-\vec{a}_1)\times\vec{b}|}{|\vec{b}|}.$$ Now, $$\vec{a}_2-\vec{a}_1 = (2,1,-1).$$

Step 3: Compute the cross product: $$(2,1,-1) \times (2,3,6) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 2 & 3 & 6 \end{vmatrix}$$ $$= 9\hat{i} - 14\hat{j} + 4\hat{k}.$$

Step 4: Compute magnitudes: $$|(\vec{a}_2-\vec{a}_1)\times\vec{b}| = \sqrt{9^2+(-14)^2+4^2} = \sqrt{293},$$ $$|\vec{b}| = \sqrt{2^2+3^2+6^2} = \sqrt{49}=7.$$

Step 5: Therefore, $$d = \frac{\sqrt{293}}{7}.$$

Answer: $\frac{\sqrt{293}}{7}$ units

Question 14 [CBSE 2023]

Find the coordinates of the point where the line passing through the points $(3, -4, -5)$ and $(2, -3, 1)$ crosses the plane $2x + y + z = 7$.

Solution: Step 1: The line passes through $$A(3,-4,-5), \quad B(2,-3,1).$$ Its direction ratios are $$(2-3,\ -3-(-4),\ 1-(-5)) = (-1,1,6).$$ Hence its equation is $$\frac{x-3}{-1} = \frac{y+4}{1} = \frac{z+5}{6} = \lambda.$$

Step 2: Write a general point on the line: $$x = -\lambda + 3, \quad y = \lambda - 4, \quad z = 6\lambda - 5.$$

Step 3: Since the point lies on the plane $2x+y+z=7$, substitute: $$2(-\lambda+3) + (\lambda-4) + (6\lambda-5) = 7.$$

Step 4: Simplify: $$-2\lambda + 6 + \lambda - 4 + 6\lambda - 5 = 7$$ $$5\lambda - 3 = 7$$ $$5\lambda = 10$$ $$\lambda = 2.$$

Step 5: Put $\lambda=2$ into the coordinates: $$x = -2+3 = 1,$$ $$y = 2-4 = -2,$$ $$z = 12-5 = 7.$$

Answer: $(1, -2, 7)$

Question 15 [CBSE 2025]

Find the equation of the plane passing through the line of intersection of the planes $x+y+z=1$ and $2x+3y+4z=5$, which is perpendicular to the plane $x-y+z=0$.

Solution: Step 1: The family of planes passing through the line of intersection of $$x+y+z-1=0 \quad \text{and} \quad 2x+3y+4z-5=0$$ is $$(x+y+z-1) + \lambda(2x+3y+4z-5) = 0.$$

Step 2: Expand and identify the normal vector of the required plane: $$(1+2\lambda)x + (1+3\lambda)y + (1+4\lambda)z - (1+5\lambda)=0.$$ So its normal vector is $$\vec{n}_1 = (1+2\lambda,\ 1+3\lambda,\ 1+4\lambda).$$

Step 3: The given plane $x-y+z=0$ has normal vector $$\vec{n}_2 = (1,-1,1).$$ Since the two planes are perpendicular, their normal vectors are perpendicular: $$\vec{n}_1\cdot\vec{n}_2 = 0.$$ Thus, $$(1+2\lambda)(1) + (1+3\lambda)(-1) + (1+4\lambda)(1) = 0.$$

Step 4: Simplify: $$1+2\lambda -1 -3\lambda +1 +4\lambda = 0$$ $$1 + 3\lambda = 0$$ $$\lambda = -\frac{1}{3}.$$

Step 5: Substitute back into the family: $$(x+y+z-1) - \frac{1}{3}(2x+3y+4z-5)=0.$$ Multiply by 3: $$3x+3y+3z-3 -2x-3y-4z+5 = 0$$ $$x - z + 2 = 0.$$

Answer: $x - z + 2 = 0$

Question 16 [CBSE 2024]

Find the distance between the parallel planes $2x - y + 3z + 4 = 0$ and $6x - 3y + 9z - 3 = 0$.

Solution: Step 1: Make the coefficients of $x,y,z$ identical. Divide the second equation by 3: $$2x - y + 3z - 1 = 0.$$ So the two planes are $$\pi_1: 2x-y+3z+4=0,$$ $$\pi_2: 2x-y+3z-1=0.$$

Step 2: For parallel planes $$Ax+By+Cz+D_1=0, \quad Ax+By+Cz+D_2=0,$$ the distance is $$d = \frac{|D_1-D_2|}{\sqrt{A^2+B^2+C^2}}.$$ Here, $$A=2, \quad B=-1, \quad C=3, \quad D_1=4, \quad D_2=-1.$$

Step 3: Substitute: $$d = \frac{|4-(-1)|}{\sqrt{2^2+(-1)^2+3^2}} = \frac{5}{\sqrt{4+1+9}} = \frac{5}{\sqrt{14}}.$$

Answer: $\frac{5}{\sqrt{14}}$ units

Question 17 [CBSE 2022]

Find the image of the point $(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$.

Solution: Step 1: Let the given point be $$P(1,6,3).$$ A general point on the line is obtained by taking the common parameter as $\lambda$: $$Q(\lambda,\ 2\lambda+1,\ 3\lambda+2).$$ If $Q$ is the foot of the perpendicular from $P$ to the line, then $\overrightarrow{PQ}$ must be perpendicular to the line direction vector $(1,2,3)$.

Step 2: Compute $$\overrightarrow{PQ} = (\lambda-1,\ 2\lambda+1-6,\ 3\lambda+2-3)$$ $$= (\lambda-1,\ 2\lambda-5,\ 3\lambda-1).$$

Step 3: Apply perpendicularity: $$(\lambda-1)(1) + (2\lambda-5)(2) + (3\lambda-1)(3) = 0.$$ So, $$\lambda -1 + 4\lambda -10 + 9\lambda -3 = 0$$ $$14\lambda -14 = 0$$ $$\lambda = 1.$$

Step 4: Therefore the foot of the perpendicular is $$Q(1,3,5).$$

Step 5: Let the image point be $P'(x',y',z')$. Since $Q$ is the midpoint of $PP'$, use midpoint formula: $$1 = \frac{1+x'}{2} \implies x'=1,$$ $$3 = \frac{6+y'}{2} \implies y'=0,$$ $$5 = \frac{3+z'}{2} \implies z'=7.$$

Answer: $(1, 0, 7)$

Question 18 [CBSE 2026]

Find the image of the point $P(1, 3, 4)$ in the plane $2x - y + z + 3 = 0$.

Solution: Step 1: The normal vector of the plane is $$(2,-1,1).$$ Hence the perpendicular line through $P(1,3,4)$ is $$\frac{x-1}{2} = \frac{y-3}{-1} = \frac{z-4}{1} = \lambda.$$ So a general point on this normal line is $$Q(2\lambda+1,\ -\lambda+3,\ \lambda+4).$$

Step 2: Since the foot of the perpendicular lies on the plane, substitute into $$2x-y+z+3=0.$$ Thus, $$2(2\lambda+1) - (-\lambda+3) + (\lambda+4) + 3 = 0.$$ Simplify: $$4\lambda+2 + \lambda-3 + \lambda+4 + 3 = 0$$ $$6\lambda + 6 = 0$$ $$\lambda = -1.$$

Step 3: Therefore the foot of the perpendicular is $$Q(-1,4,3).$$

Step 4: If $P'(x',y',z')$ is the image of $P$, then $Q$ is the midpoint of $PP'$: $$-1 = \frac{1+x'}{2} \implies x'=-3,$$ $$4 = \frac{3+y'}{2} \implies y'=5,$$ $$3 = \frac{4+z'}{2} \implies z'=2.$$

Answer: $(-3, 5, 2)$

Question 19 [CBSE 2023]

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $2x - 3y + 4z - 6 = 0$.

Solution: Step 1: The normal vector of the plane is $$(2,-3,4).$$ So the perpendicular from the origin to the plane lies along the line $$\frac{x}{2} = \frac{y}{-3} = \frac{z}{4} = \lambda.$$ Thus a general point on this line is $$Q(2\lambda,\ -3\lambda,\ 4\lambda).$$

Step 2: Since $Q$ lies on the plane, $$2(2\lambda) - 3(-3\lambda) + 4(4\lambda) - 6 = 0.$$ So, $$4\lambda + 9\lambda + 16\lambda - 6 = 0$$ $$29\lambda = 6$$ $$\lambda = \frac{6}{29}.$$

Step 3: Substitute back: $$Q = \left(\frac{12}{29},\ -\frac{18}{29},\ \frac{24}{29}\right).$$

Answer: $(12/29, -18/29, 24/29)$

Question 20 [CBSE 2025]

Find the equation of the plane containing the line $\frac{x+1}{-3} = \frac{y-3}{2} = \frac{z+2}{1}$ and the point $B(0, 7, -7)$.

Solution: Step 1: A point on the line is obtained by taking the common parameter as 0: $$A(-1,3,-2).$$ The direction vector of the line is $$\vec{b} = (-3,2,1).$$

Step 2: Find the vector joining the point on the line to the external point: $$\overrightarrow{AB} = (0-(-1),\ 7-3,\ -7-(-2)) = (1,4,-5).$$

Step 3: The plane contains both the line direction vector and $\overrightarrow{AB}$, so a normal vector is $$\vec{n} = \overrightarrow{AB} \times \vec{b}.$$ Thus, $$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & -5 \\ -3 & 2 & 1 \end{vmatrix}$$ $$= 14\hat{i} + 14\hat{j} + 14\hat{k}.$$ A simplified normal vector is $$(1,1,1).$$

Step 4: Use point-normal form through $B(0,7,-7)$: $$1(x-0) + 1(y-7) + 1(z+7) = 0.$$ Simplifying, $$x+y+z=0.$$

Answer: $x + y + z = 0$

Question 21 [CBSE 2024]

Find the value of $k$ if the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{4}$ lies in the plane $x + 2y + kz = -1$.

Solution: Step 1: For a line to lie in a plane, two conditions are needed:

1. A point on the line must satisfy the plane equation.
2. The direction vector of the line must be perpendicular to the normal of the plane.

Step 2: A point on the line is $$(1,-1,3).$$ Substitute into the plane equation: $$1 + 2(-1) + k(3) = -1.$$ So, $$1 - 2 + 3k = -1$$ $$-1 + 3k = -1$$ $$3k = 0 \implies k=0.$$

Step 3: Verify the direction condition. The line has direction vector $$(2,-1,4),$$ and the plane normal is $$(1,2,k).$$ For $k=0$, $$(2,-1,4)\cdot(1,2,0) = 2 - 2 + 0 = 0.$$ So the line is parallel to the plane and also passes through a point of the plane. Hence it lies in the plane.

Answer: $k = 0$

Question 22 [CBSE 2026]

Find the vector equation of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines $\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}$ and $\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}$.

Solution: Step 1: The direction ratios of the two given lines are $$\vec{b}_1 = (3,-16,7), \quad \vec{b}_2 = (3,8,-5).$$ A line perpendicular to both must have a direction vector perpendicular to both $\vec{b}_1$ and $\vec{b}_2$. Hence its direction vector is their cross product.

Step 2: Compute $$\vec{b} = \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}.$$ Expanding, $$\vec{b} = \hat{i}(80-56) - \hat{j}(-15-21) + \hat{k}(24-(-48))$$ $$= 24\hat{i} + 36\hat{j} + 72\hat{k}.$$ A simplified direction vector is $$2\hat{i} + 3\hat{j} + 6\hat{k}.$$

Step 3: The required line passes through $(1,2,-4)$, whose position vector is $$\vec{a} = \hat{i} + 2\hat{j} - 4\hat{k}.$$ So its vector equation is $$\vec{r} = \vec{a} + \lambda\vec{b}.$$ Therefore, $$\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}).$$

Answer: $\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$

Question 23 [CBSE 2023]

Find the distance of the point $P(-1, -5, -10)$ from the point of intersection of the line $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}$ and the plane $x - y + z = 5$.

Solution: Step 1: A general point on the line is $$Q(3\lambda+2,\ 4\lambda-1,\ 12\lambda+2).$$

Step 2: Since the point of intersection lies on the plane $x-y+z=5$, substitute: $$(3\lambda+2) - (4\lambda-1) + (12\lambda+2) = 5.$$ Simplify: $$3\lambda+2 -4\lambda+1 +12\lambda+2 = 5$$ $$11\lambda + 5 = 5$$ $$\lambda = 0.$$

Step 3: So the intersection point is $$Q(2,-1,2).$$

Step 4: Find the distance from $P(-1,-5,-10)$ to $Q(2,-1,2)$: $$PQ = \sqrt{(2+1)^2 + (-1+5)^2 + (2+10)^2}$$ $$= \sqrt{3^2 + 4^2 + 12^2} = \sqrt{169} = 13.$$

Answer: 13 units