Let l 1 be the line in x y -plane with x and y intercepts 1 8 and 1 4 2 respectively, and l 2 be the line in…

Given,

Line $L_1:\frac{x-\frac{1}{8}}{\frac{1}{8}}=\frac{y}{-\frac{1}{4\sqrt{2}}}=\frac{z}{0}$

or $\frac{x-\frac{1}{8}}{1}=\frac{y}{-\sqrt{2}}=\frac{z}{0} ...\left(1\right)$

Equation of $L_2:\frac{x+\frac{1}{8}}{-6\sqrt{3}}=\frac{y}{0}=\frac{z}{8}$

Shortest distance between two lines is given by,

$d=\left|\frac{\left(\overrightarrow{c}-\overrightarrow{a}\right)·\left(\overrightarrow{b}\times \overrightarrow{d}\right)}{\left|\overrightarrow{b}\times \overrightarrow{d}\right|}\right|=\frac{\left(\frac{1}{4}\hat{i}\right)·\left(4\sqrt{2}\hat{i}+4\hat{j}+3\sqrt{6}\hat{k}\right)}{\sqrt{{\left(4\sqrt{2}\right)}^2+4^2+{\left(3\sqrt{6}\right)}^2}}$

$=\frac{\sqrt{2}}{\sqrt{32+16+54}}=\frac{1}{\sqrt{51}}$

$\therefore d^{-2}=51$