If A=(1, –2, –1), B=(4, 0, –3), C=(1, 2, –1), D=(2, –4, –5), then distance between AB and CD is

$\begin{array}{rcl}\overline{\mathrm{a}}& =& \overline{\mathrm{OA}}=(1,-2,-1) , \overline{\mathrm{b}}=\overline{\mathrm{AB}}=(3,2,-2)\\ \overline{\mathrm{c}}& =& \overline{\mathrm{OC}}=(1,2,-1) , \overline{\mathrm{d}}=\overline{\mathrm{CD}}=(1,-6,-4)\end{array}$

$\begin{array}{rcl}\overline{\mathrm{a}}-\overline{\mathrm{c}}& =& (0,-4,0) \overline{\mathrm{b}}\times \overline{\mathrm{d}}=\left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 3& 2& -2\\ 1& -6& -4\end{array}\right|\end{array}$

$\begin{array}{rcl} & =& \mathrm{i}(-20)-\mathrm{j}(-10)+\mathrm{k}(-20)\\ & =& 10(2\mathrm{i}+\mathrm{j}-2\mathrm{k})\end{array}$

$\begin{array}{rcl}\mathrm{S}.\mathrm{D}& =& \left|\frac{10(-4)}{10\times 3}\right|=\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\end{array}$