JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Three Dimensional Geometry

The line passes through $(2,-1,-3)$ with direction ratios $(3,-2,-1)$. Since the point lies on the plane: $2p+q-3=5$, so $2p+q=8$. Since the line lies on the plane, its direction vector is perpendicular to the normal $(p,-q,1)$: $3p+2q-1=0$. Solving the system: from the second equation $q=\frac{1-3p}{2}$. Substituting into the first: $2p+\frac{1-3p}{2}=8$ gives $p=15$, hence $q=-22$. The plane is $15x+22y+z=5$. Distance from origin $=\frac{5}{\sqrt{15^2+22^2+1}}=\frac{5}{\sqrt{710}}=\sqrt{\frac{25}{710}}=\sqrt{\frac{5}{142}}$.