QJEEM25Ma8172: JEE Main 2025 Mathematics - Class 12 JEEM25Ma8172 Solution: Let P be the foot of the perpendicular from the point on the…

$\begin{aligned}& \mathrm{L}: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+1}{-1}=\frac{\mathrm{z}-2}{2}=\mu \\& \mathrm{P}(\mu+1,-\mu-1,2 \mu+2) \\& \overrightarrow{\mathrm{AP}} \cdot \overrightarrow{\mathrm{~d}}=0 \Rightarrow(\mu,-\mu-3,2 \mu) \cdot(1,-1,2)=0 \\& \Rightarrow \mu+\mu+3+4 \mu=0 \Rightarrow \mu=-\frac{1}{2} \\& \therefore \mathrm{P}\left(\frac{-1}{2}+1,+\frac{1}{2}-1,2\left(\frac{-1}{2}\right)+2\right) \\& \mathrm{P}\left(\frac{1}{2}, \frac{-1}{2}, 1\right)\end{aligned}$

$\begin{array}{c|c|c}\mu+1=-1+\lambda & -\mu-1=1-\lambda & 2 \mu+2=-2+\lambda \\\mu=\lambda-2 & \mu=\lambda-2 & \downarrow\end{array}$

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

$\begin{aligned}& \therefore \quad \mu=-2, \lambda=0 \\& \therefore \quad \mathrm{Q} \equiv(-1,1-2) \\& \mathrm{P}\left(\frac{1}{2}, \frac{-1}{2}, 1\right) \text { and } \mathrm{Q}(-1,1,-2) \\& \mathrm{PQ}=\sqrt{\left(\frac{1}{2}+1\right)^2+\left(\frac{-1}{2}-1\right)^2+(1+2)^2} \\& =\sqrt{\frac{9}{4}+\frac{9}{4}+9}=\sqrt{\frac{54}{4}} \\& \therefore \quad 2(\mathrm{PQ})^2=2\left(\frac{54}{4}\right)=27\end{aligned}$