JEE Main 2025 — 29 January, Evening Shift — Question 54

$\mathrm{L}: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+1}{-1}=\frac{\mathrm{z}-2}{2}=\mu$

$\mathrm{P}(\mu+1,-\mu-1,2 \mu+2)$

$\overrightarrow{\mathrm{AP}} \cdot \overrightarrow{\mathrm{d}}=0 \Rightarrow(\mu,-\mu-3,2 \mu) \cdot(1,-1,2)=0$

$\Rightarrow \mu+\mu+3+4 \mu=0 \Rightarrow \mu=-\frac{1}{2}$

$\therefore \mathrm{P}\left(\frac{-1}{2}+1,+\frac{1}{2}-1,2\left(\frac{-1}{2}\right)+2\right)$

$\mathrm{P}\left(\frac{1}{2}, \frac{-1}{2}, 1\right)$

Now general pt. on $\mathrm{L}_{2}$ is $\mathrm{Q}(-1+\lambda, 1-\lambda,-2+\lambda)$

Equate it with general pt of L

$\mu+1=-1+\lambda$ & $-\mu-1=1-\lambda$ & $2 \mu+2=-2+\lambda$

$\mu=\lambda-2$ & $\mu=\lambda-2$ & $\downarrow$

$$\begin{aligned} & 2(\lambda-2)+2=-2+\lambda \\ & 2 \lambda-4+2=-2+\lambda \end{aligned}$$

$\therefore \mu=-2, \lambda=0$

$\therefore \mathrm{Q} \equiv(-1,1-2)$

$\mathrm{P}\left(\frac{1}{2}, \frac{-1}{2}, 1\right)$ and $\mathrm{Q}(-1,1,-2)$

$\mathrm{PQ}=\sqrt{\left(\frac{1}{2}+1\right)^{2}+\left(\frac{-1}{2}-1\right)^{2}+(1+2)^{2}}$

$$=\sqrt{\frac{9}{4}+\frac{9}{4}+9}=\sqrt{\frac{54}{4}}$$

$\therefore 2(\mathrm{PQ})^{2}=2\left(\frac{54}{4}\right)=27$