JEE Main 2025 (Online) 7th April Evening ShiftQuestion 1
Consider the lines $L_1:x-1=y-2=z$ and $L_2:x-2=y=z-1$. Let the feet of the perpendiculars from the point $P(5,1,-3)$ on the lines $L_1$ and $L_2$ be $Q$ and $R$ respectively. If the area of the triangle $PQR$ is $A$, then $4A^{2}$ is equal to :
3D Geometry
Solution
Feet: $Q=(1,2,0)$ (parameter $0$ on $L_1$), $R=(2,0,1)$. $\vec{PQ}=(-4,1,3)$, $\vec{PR}=(-3,-1,4)$, cross product $=(7,7,7)$. $A=\frac12\sqrt{147}$, so $4A^2=147$.