Find the shortest distance between the lines
$\vec r = (\hat i + 2\hat j + \hat k) + \lambda (\hat i - \hat j + \hat k)$ and
$\vec r = 2\hat i - \hat j - \hat k + \mu (2\hat i + \hat j + 2\hat k)$

. : Comparing the given lines by $\vec b = {\vec a_1} + \lambda {\vec b_1}$ and $\vec r = {\vec a_2} + \mu {\vec b_2}$,

we get ${\vec a_1} = \hat i + 2\hat j + \hat k,{\vec a_2} = 2\hat i - \hat j - \hat k$

and ${\vec b_1} = \hat i - \hat j + \hat k,{\vec b_2} = 2\hat i + \hat j + 2\hat k$

Now, ${\vec a_1} - {\vec a_2} = (2\hat i - \hat j - \hat k) - (\hat i + 2\hat j + \hat k) = \hat i - 3\hat j - 2\hat k$

and ${\vec b_1} \times {\vec b_2} = \left| {\begin{array}{cccccccccccccccccccc}{\hat i}&{\hat j}&{\hat k}\\1&{ - 1}&1\\2&1&2\end{array}} \right|$

$= \hat i( - 2 - 1) - \hat j(2 - 2) + \hat k(1 + 2) = - 3\hat i + 3\hat k$

Required shortest distance
$= \left| {\cfrac{{({{\vec b}_1} \times {{\vec b}_2}) \cdot ({{\vec a}_2} - {{\vec a}_1})}}{{|{{\vec b}_1} \times {{\vec b}_2}|}}} \right|$

….(i)
$= \left| {\cfrac{{( - 3\hat i + 3\hat k) \cdot (\hat i - 3\hat j - 2\hat k)}}{{| - 3\hat i + 3\hat k|}}} \right|$

$= \left| {\cfrac{{( - 3) \times 1 + 0 \times ( - 3) + 3 \times ( - 2)}}{{\sqrt {{{( - 3)}^2} + {3^2}} }}} \right|$

$= \cfrac{9}{{\sqrt {18} }} = \cfrac{9}{{3\sqrt 2 }} = \cfrac{3}{{\sqrt 2 }}$ units