Question

Let $l_{1}$ be the line in $xy$-plane with $x$ and $y$ intercepts $\frac{1}{8}$ and $\frac{1}{4 \sqrt{2}}$ respectively, and $l_{2}$ be the line in $zx$-plane with $x$ and $z$ intercepts $-\frac{1}{8}$ and $-\frac{1}{6 \sqrt{3}}$ respectively. If $d$ is the shortest distance between the line $l_{1}$ and $l_{2}$, then $d ^{-2}$ is equal to

Answer 1

$8 x+4 \sqrt{2} y=1, z=0$
$\Rightarrow \frac{x-\frac{1}{8}}{1}=\frac{y-0}{-\sqrt{2}}=\frac{z-0}{0}=\lambda$
$-8 x-6 \sqrt{3} z=1, y=0$
$\Rightarrow \frac{x+\frac{1}{8}}{3 \sqrt{3}}=\frac{y-0}{0}=\frac{z-0}{-4}$
$\begin{vmatrix}\frac{1}{4} & 0 & 0 \\ 1 & -\sqrt{2} & 0 \\ 3 \sqrt{3} & 0 & -4\end{vmatrix}=\sqrt{2}$
$d =\frac{1}{\sqrt{51}}$
$\frac{1}{ d ^{2}}=51$