Let P be the foot of the perpendicular from the point (1,2,2) on...

General point on line $\mathrm{L}: \frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-2}{2}$ is $(\lambda+1,-\lambda-1,2 \lambda+2)$
DR's of PM are $(\lambda,-\lambda-3,2 \lambda)$ $P M \perp L$ $$\begin{aligned} & \Rightarrow \lambda+(-1)(-\lambda-3)+2(2 \lambda)=0 \\ & \Rightarrow 6 \lambda+3=0 \end{aligned}$$
P\left(\frac{1}{2}, \frac{-1}{2}, 1\right)$ Let another line $L^{\prime}: \frac{x+1}{1}=\frac{y-1}{-1}=\frac{z+2}{1}$ General point on line $L^{\prime}$ is $(\mu-1,-\mu+1, \mu-2)$ Point of intersection of line $L$ and $L^{\prime}$ is\begin{array}{l|l} \lambda+1=\mu-1 & 2 \lambda+2=\mu-2 \\ \Rightarrow \mu-\lambda=2 \ldots(1) & \Rightarrow 2 \lambda=\mu-4 \end{array}$$<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/tv00fIUHl_p5piJ_sZzeDfj2UVox06HIZYHs0AXJfoI.original.fullsize.png"><br>$$\begin{aligned} & Q(-1,1,-2) \\ & 2(P Q)^2=2\left(\left(\frac{1}{2}+1\right)^2+\left(\frac{-1}{2}-1\right)^2+(1+2)^2\right) \\ & =2\left(\frac{9}{4}+\frac{9}{4}+9\right) \\ & =27\end{aligned}$$