JEE Main 2023 (Online) 31st January Morning ShiftQuestion 1
Let the line $L:\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}$ intersect the plane $2x+y+3z=16$ at the point $P$. Let the point $Q$ be the foot of perpendicular from the point $R(1,-1,-3)$ on the line $L$. If $\alpha$ is the area of triangle $PQR$, then $\alpha^{2}$ is equal to ______.
3D Geometry
Answer$180$
Solution
Plane gives $t=1$, $P=(3,-2,4)$. Foot of perpendicular from $R$: $t=-1$, $Q=(-1,0,2)$. $\vec{QP}=(4,-2,2)$, $\vec{QR}=(2,-1,-5)$ are perpendicular, so $\alpha=\frac12\sqrt{24}\sqrt{30}$ and $\alpha^2=180$.