JEE Main 2026 — 6 April, Evening Shift — Question 43

$\mathrm{P}(0,-5,0)$

$$\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}}{1}=\frac{\mathrm{z}+1}{-2}=\mathrm{k}$$

Let $\mathrm{R}\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ $2 \mathrm{k}+1, \mathrm{k},-2 \mathrm{k}-1$ $(2 \mathrm{k}+1)(2)+(\mathrm{k}+5)(1)+(-2 \mathrm{k}-1)(-2)=0$ $\Rightarrow 9 \mathrm{k}+9=0 \Rightarrow \mathrm{k}=-1$ $\therefore \mathrm{M}_{1} \Rightarrow(-1,-1,1)$

$$R \Rightarrow(-2,3,2)$$

Q(0, -1/2, 0) $\frac{x-1}{-1}=\frac{y+9}{4}=\frac{z+1}{1}=\lambda$ $(-\lambda+1,4 \lambda-9, \lambda-1)$ $\mathrm{S}\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right)$ $(-\lambda+1)(-1)+\left(4 \lambda-\frac{17}{2}\right)(4)+(\lambda-1)(1)=0$ $\Rightarrow 18 \lambda-36=0 \Rightarrow \lambda=2$ $\mathrm{M}_{2} \Rightarrow(-1,-1,1)$ $\mathrm{S} \Rightarrow(-2,-3 / 2,2)$

$$\begin{aligned} & \mathrm{PR}=\sqrt{4+64+4}=\sqrt{72}=6 \sqrt{2} \\& \mathrm{QS}=\sqrt{4+1+4}=\sqrt{9}=3 \\& \text { Ar. }=\frac{1}{2} \cdot 6 \sqrt{2} \cdot 3=9 \sqrt{2} \\& (\text { Ar. })^{2}=(9 \sqrt{2})^{2}=162 \text { Sq.unit } \end{aligned}$$