Shortest distance between the lines (x - 1)/(2) = (y + 8)/(- 7) =...

Given lines are

$\dfrac{x - 1}{2} = \dfrac{y + 8}{- 7} = \dfrac{z - 4}{5}$ and

$\dfrac{x - 1}{2} = \dfrac{y - 2}{1} = \dfrac{z - 6}{- 3}$

Corresponding vectors are:

\overset{\to}{a} = \overset{^}{i} - 8\overset{^}{j} + 4\overset{^}{k}, \overset{\to}{p} = 2\overset{^}{i} - 7\overset{^}{j} + 5\overset{^}{k}

\overset{\to}{b} = \overset{^}{i} + 2\overset{^}{j} + 6\overset{^}{k}, \overset{\to}{q} = 2\overset{^}{i} + \overset{^}{j} - 3\overset{^}{k}

$S . D . = \dfrac{| \overset{\to}{a} - \overset{\to}{b}\overset{\to}{p}\overset{\to}{q} |}{| \overset{\to}{p} \times \overset{\to}{q} |}$

\overset{\to}{a} - \overset{\to}{b} = - 10\overset{^}{j} - 2\overset{^}{k}

[ \overset{\to}{a} - \overset{\to}{b}\overset{\to}{p}\overset{\to}{q} ] = \left|\begin{cases}0 & - 10 & - 2 \\ 2 & - 7 & 5 \\ 2 & 1 & - 3\end{cases}\right|

$= 0 - ( - 10 ) ( - 6 - 10 ) + ( - 2 ) ( 2 + 14 )$

$= 0 - 160 - 32 = - 192$

$| [ \overset{\to}{a} - \overset{\to}{b}\overset{\to}{p}\overset{\to}{q} ] | = 192$

\overset{\to}{p} \times \overset{\to}{q} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 2 & - 7 & 5 \\ 2 & 1 & - 3\end{cases}\right|

= \overset{^}{i}\left(16\right) - \overset{^}{j}\left(- 16\right) + \overset{^}{k}\left(16\right)

= 16\left(\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right)

$| \overset{\to}{p} \times \overset{\to}{q} | = 16\sqrt{3}$

$S . D . = \dfrac{| \overset{\to}{a} - \overset{\to}{b}\overset{\to}{p}\overset{\to}{q} |}{| \overset{\to}{p} \times \overset{\to}{q} |} = \dfrac{192}{16\sqrt{3}} = \dfrac{12}{\sqrt{3}} = 4\sqrt{3}$