Shortest distance between the lines x - 1 2 = y + 8 - 7 = z - 4 5  and  x - 1 2 = y - 2 1 = z - 6…

Given lines are

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

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

Corresponding vectors are:

$\overrightarrow{a}=\hat{i}-8\hat{j}+4\hat{k}$, $\overrightarrow{p}=2\hat{i}-7\hat{j}+5\hat{k}$

$\overrightarrow{b}=\hat{i}+2\hat{j}+6\hat{k}$, $\overrightarrow{q}=2\hat{i}+\hat{j}-3\hat{k}$

$S.D.=\frac{|\overrightarrow{a}-\overrightarrow{b} \overrightarrow{p} \overrightarrow{q} |}{|\overrightarrow{p}\times \overrightarrow{q}|}$

$\overrightarrow{a}-\overrightarrow{b}=-10\hat{j}-2\hat{k}$

$[\overrightarrow{a}-\overrightarrow{b} \overrightarrow{p} \overrightarrow{q}]=\left|\begin{array}{ccc}0& -10& -2\\ 2& -7& 5\\ 2& 1& -3\end{array}\right|$

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

$=0-160-32=-192$

$\left|\right[\overrightarrow{a}-\overrightarrow{b} \overrightarrow{p} \overrightarrow{q}\left]\right|=192$

$\overrightarrow{p}\times \overrightarrow{q}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -7& 5\\ 2& 1& -3\end{array}\right|$

$=\hat{i}\left(16\right)-\hat{j}\left(-16\right)+\hat{k}\left(16\right)$

$=16\left(\hat{i}+\hat{j}+\hat{k}\right)$

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

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