JEE Main 2026 — Three Dimensional Geometry Question with Solution

Line L passes through P(1,1,1) perpendicular to two lines with directions $(4,1,1)$ and $(1,1,0)$.
Direction of L: $(4,1,1)\times (1,1,0)=(-1,1,3)$
Line L: $(x,y,z)=(1,1,1)+t(-1,1,3)$
Q is intersection with yz-plane (x=0): Setting $1-t=0$ gives $t=1$, so Q = (0, 2, 4)
Line through S(1,0,-1) parallel to L: $(x,y,z)=(1,0,-1)+s(-1,1,3)$
R is intersection with yz-plane: Setting $1-s=0$ gives $s=1$, so R = (0, 1, 2)
For parallelogram PQRS: $\overrightarrow{PQ}=(-1,1,3)$ and $\overrightarrow{PS}=(0,-1,-2)$
Area = $|\overrightarrow{PQ}\times \overrightarrow{PS}|=|(1,-2,1)|=\sqrt{1+4+1}=\sqrt{6}$
Square of area = 6