JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Three Dimensional Geometry

The line $L$ is $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}=\lambda$, so a general point on it is $(1+2\lambda ,-1+3\lambda ,4\lambda )$.

Since $R(5,p,q)$ lies on $L$: $1+2\lambda =5\Rightarrow \lambda =2$, giving $R=(5,5,8)$.

Let $T$ be the foot of the perpendicular from $Q(7,-2,5)$ to $L$: $T=(1+2\lambda ,-1+3\lambda ,4\lambda )$ for some $\lambda$. Then $\overrightarrow{QT}=(2\lambda -6,3\lambda +1,4\lambda -5)$ must be perpendicular to the direction vector $(2,3,4)$ of $L$:

so $T=(3,2,4)$.

Since $P$ is the reflection (image) of $Q$ in the line $L$, $T$ is the midpoint of $PQ$:


Because $QT\perp L$ and $R$ lies on $L$, triangle $QTR$ is right-angled at $T$; the same holds for $PTR$ since $P$ is the mirror image of $Q$ through $T$. So $PQ$ is perpendicular to line $L$ at $T$, and the perpendicular distance from $R$ (on $L$) to line $PQ$ is simply $RT$.

Compute:



So


and the square of the area is