JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Three Dimensional Geometry

The first line lies in the $xy$-plane with intercepts $\frac{1}{8}$ and $\frac{1}{4\sqrt{2}}$, so it is $8x+4\sqrt{2}y=1,z=0$, through $(\frac{1}{8},0,0)$ with direction $(1,-\sqrt{2},0)$. The second lies in the $zx$-plane with intercepts $-\frac{1}{8}$ and $-\frac{1}{6\sqrt{3}}$, giving $8x+6\sqrt{3}z=-1,y=0$, through $(-\frac{1}{8},0,0)$ with direction $(3\sqrt{3},0,-4)$. Their cross product is $(1,-\sqrt{2},0)\times (3\sqrt{3},0,-4)=(4\sqrt{2},4,3\sqrt{6})$, whose squared length is $32+16+54=102$. The vector joining the chosen points is $(-\frac{1}{4},0,0)$, and its scalar product with the cross product is $-\frac{1}{4}\cdot 4\sqrt{2}=-\sqrt{2}$. The shortest distance is $d=\frac{|-\sqrt{2}|}{\sqrt{102}}$, so $d^2=\frac{2}{102}=\frac{1}{51}$ and $d^{-2}=51$.