JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Three Dimensional Geometry

The line in parametric form is $(2\lambda +1,-\lambda -1,\lambda +3)$. Substituting into $2x+y+3z=16$ gives $6\lambda +10=16$, so $\lambda =1$ and $P=(3,-2,4)$. For the foot of the perpendicular from $R(1,-1,-3)$, requiring $\overrightarrow{QR}$ perpendicular to the direction $(2,-1,1)$ gives $6\lambda +6=0$, so $\lambda =-1$ and $Q=(-1,0,2)$. Since $Q$ is that foot, the triangle is right-angled at $Q$, with $\overrightarrow{QR}=(2,-1,-5)$ and $\overrightarrow{QP}=(4,-2,2)$. Their cross product is $(-12,-24,0)$ of magnitude $\sqrt{720}$, so the area is $\frac{1}{2}\sqrt{720}$ and ${\alpha }^2=\frac{720}{4}=180$.