Let l₁ be the line in xy-plane with x and y intercepts ${1 \over 8}$ and ${1 \over {4\sqrt 2 }}$ respectively, and l₂ be the line in zx-plane with x and z intercepts $- {1 \over 8}$ and $- {1 \over {6\sqrt 3 }}$ respectively. If d is the shortest distance between the line l₁ and l₂, then d^$-$2 is equal to _______________.

<p>$${{x - {1 \over 8}} \over {{1 \over 8}}} = {y \over { - {1 \over {4\sqrt 2 }}}} = {z \over 0}\,\,\,\,\,\,\,\,\,\,\,\,\,$$ ______L₁</p> <p>or ${{x - {1 \over 8}} \over 1} = {y \over { - \sqrt 2 }} = {z \over 0}$ ..... (i)</p> <p>Equation of L₂</p> <p>${{x + {1 \over 8}} \over { - 6\sqrt 3 }} = {y \over 0} = {z \over 8}$ ...... (ii)</p> <p>$$d = \left| {{{(\overrightarrow c - \overrightarrow a )\,.\,(\overrightarrow b \times \overrightarrow d )} \over {\left| {\overrightarrow b \times \overrightarrow d } \right|}}} \right|$$</p> <p>$$ = {{\left( {{1 \over 4}\widehat i} \right)\,.\,\left( {4\sqrt 2 \widehat i + 4\widehat j + 3\sqrt 6 \widehat k} \right)} \over {\sqrt {{{\left( {4\sqrt 2 } \right)}^2} + {4^2} + {{\left( {3\sqrt 6 } \right)}^2}} }}$$</p> <p>$= {{\sqrt 2 } \over {\sqrt {32 + 16 + 54} }} = {1 \over {\sqrt {51} }}$</p> <p>${d^{ - 2}} = 51$</p>