The shortest distance between the lines $\frac{x+7}{-6}=\frac{y-6}{7}=z$ and $\frac{7-x}{2}=y-2=z-6$ is :

<p>${L_1}:{{x + 7} \over 6} = {{y - 6} \over 7} = {{z - 0} \over 1}$</p> <p>Any point on it ${\overrightarrow a _1}( - 7,6,0)$ and L₁ is parallel to ${\overrightarrow b _1}( - 6,7,1)$</p> <p>${L_2}:{{x - 7} \over { - 2}} = {{y - 2} \over 1} = {{z - 6} \over 1}$</p> <p>Any point on it ${\overrightarrow a _2}(7,2,6)$ and L₂ is parallel to ${\overrightarrow b _2}( - 2,1,1)$</p> <p>Shortest distance between L₁ and L₂</p> <p>$$ = \left| {{{({{\overrightarrow a }_2} - {{\overrightarrow a }_1})\,.\,({{\overrightarrow b }_1} \times {{\overrightarrow b }_2})} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}}} \right| = \left| {{{( - 14,4, - 6)\,.\,(3,2,4)} \over {\sqrt {9 + 4 + 16} }}} \right|$$</p> <p>$= 2\sqrt {29}$.</p>