JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Three Dimensional Geometry

The foot of the perpendicular from $P(0,-5,0)$ to the first line, using a general point $(1+2k,k,-1-2k)$, is found from $9k+9=0$, so $k=-1$, the foot is $(-1,-1,1)$ and the image is $R(-2,3,2)$. For $Q(0,-\frac{1}{2},0)$ and the second line, a general point $(1-\lambda ,-9+4\lambda ,-1+\lambda )$ gives $18\lambda -36=0$, so $\lambda =2$, the foot is $(-1,-1,1)$ and the image is $S(-2,-\frac{3}{2},2)$. Both diagonals $PR$ and $QS$ share the midpoint $(-1,-1,1)$, so $PQRS$ is a parallelogram and its area is $\frac{1}{2}|\overrightarrow{PR}\times \overrightarrow{QS}|$. With $\overrightarrow{PR}=(-2,8,2)$ and $\overrightarrow{QS}=(-2,-1,2)$, $\overrightarrow{PR}\times \overrightarrow{QS}=(18,0,18)$, of magnitude $18\sqrt{2}$, so the area is $9\sqrt{2}$. The square of the area is $162$.