JEE Main 2025 (Online) 24th January Evening ShiftQuestion 1
Let $P$ be the image of the point $Q(7,-2,5)$ in the line $L:\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z}{4}$ and $R(5,p,q)$ be a point on $L$. Then the square of the area of $\triangle PQR$ is $\_\_\_\_\_\_\_\_$ .
3D Geometry
Answer$957$
Solution
Foot of $Q$ on $L$: $t=1$, $F=(3,2,4)$, so $P=(-1,6,3)$; $R=(5,5,8)$ ($t=2$). $\overrightarrow{QP}\times\overrightarrow{QR}=(-8,8,-2)\times(-2,7,3)=(38,28,-40)$, so $\text{area}^2=\frac{1444+784+1600}4=957$.