Let $\overline{\mathrm{a}}=\overline{\mathrm{i}}+2 \overline{\mathrm{j}}-\overline{\mathrm{k}}$ and $\overline{\mathrm{b}}=\overline{\mathrm{i}}+\overline{\mathrm{j}}+\overline{\mathrm{k}}$. If $\overline{\mathrm{p}}$ is a unit vector such that $[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{p}}]$ is maximum, then $\overline{\mathrm{p}}$ =

$\begin{array}{rcl}[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{p}}]& =& (\overline{\mathrm{a}}\times \overline{\mathrm{b}})·\overline{\mathrm{p}}\le |\overline{\mathrm{a}}\times \overline{\mathrm{b}}| \left|\overline{\mathrm{p}}\right|\le |\overline{\mathrm{a}}\times \overline{\mathrm{b}}|\\ \overline{\mathrm{a}}\times \overline{\mathrm{b}}& =& \left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 1& 2& -1\\ 1& 1& 1\end{array}\right|=\mathrm{i}\left(3\right)-\mathrm{j}\left(2\right)+\mathrm{k}(-1)\\ |\overline{\mathrm{a}}\times \overline{\mathrm{b}}|& =& \sqrt{9+4+1}=\sqrt{14}\\ [\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{p}}]& \le & \sqrt{14}\\ \therefore \overline{\mathrm{p}}& =& \frac{3\mathrm{i}-2\mathrm{j}-\mathrm{k}}{\sqrt{14}}\end{array}$