Find the shortest distance between the skew lines $\overline{r}=(6\hat{i}+2\hat{j}+2\hat{k})+t (\hat{i}-2\hat{j}+2\hat{k})$ and $\overline{r}=(-4\hat{i}-\hat{k})+s(3\hat{i}-2\hat{j}-2\hat{k})$

Given lines are 

$\overline{r}=(6\overline{i}+2\overline{j}+2\overline{k})+t(\overline{i}-2\overline{j}+2\overline{k})............\left(1\right)$

and $\overline{r}=(-4\overline{i}-\overline{k})+s(3\overline{i}-2\overline{j}-2\overline{k})...........\left(2\right)$

Eq.(1) passing through the point $\overline{a}=6\overline{i}+2\overline{j}+2\overline{k}$ and parallel to the vector $\overline{b}=\overline{i}-2\overline{j}+2\overline{k}$ and 

Eq. (2) passing through the point $\overline{c}=-4\overline{i}-\overline{k}$ and parallel to the vector $\overline{d}=3\overline{i}-2\overline{j}-2\overline{k}$ Now $\overline{a}-\overline{c}=(6\overline{i}+2\overline{j}+2\overline{k})-(-4\overline{i}-\overline{k})$ 

$=10\overline{i}+2\overline{j}+3\overline{k}$ and

$\overline{b}\times \overline{d}=\left|\begin{array}{ccc}\overline{i}& \overline{j}& \overline{k}\\ 1& -2& 2\\ 3& -2& -2\end{array}\right|$

$=\overline{i}[4-(-4\left)\right]-\overline{j}[-2-(6\left)\right]+\overline{k}[-2-(-6\left)\right]$

$=8\overline{i}+8\overline{j}+4\overline{k}$

$|\overline{b}\times \overline{d}|=\sqrt{(8{)}^2+(8{)}^2+(4{)}^2}$

$=\sqrt{64+64+16}=\sqrt{144}=12$

$[\overline{a}-\overline{c}\overline{b}\overline{d}]=\left|\begin{array}{ccc}10& 2& 3\\ 1& -2& 2\\ 3& -2& -2\end{array}\right|$

$=10[4-(-4\left)\right]-2[-2-(6\left)\right]+3[-2-(-6]$

$=10\left(8\right)-2(-8)+3\left(4\right)$=108

Shortest distance between the skew lines 

$\frac{\left|\right[\overline{a}-\overline{c}\overline{b}\overline{d}\left]\right|}{|\overline{b}\times \overline{d}|}=\frac{\left|108\right|}{12}=\frac{108}{12}=9\text{ units }$