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 $\left[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{p}}\right]$ is maximum, then $\overline{\mathrm{p}}=$

$$\begin{gathered} \left[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{p}}\right] \max \\ \Rightarrow \left|\overline{\mathrm{p}}\right|=1 ,\left( \overline{\mathrm{a}}\times \overline{\mathrm{b}}, \overline{\mathrm{p}}\right)=0^{\circ } \\ \Rightarrow \overline{\mathrm{p}} \parallel \overline{\mathrm{a}}\times \overline{\mathrm{b}} \\ \overline{\mathrm{a}}\times \overline{\mathrm{b}}=\left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 1& 2& -1\\ 1& 1& 1\end{array}\right|=\mathrm{i}\left(3\right)-\mathrm{j}\left(2\right)+\mathrm{k}(-1) \\ =3\mathrm{i}-2\mathrm{j}-\mathrm{k} \\ \overline{\mathrm{p}}=\frac{3\mathrm{i}-2\mathrm{j}-\mathrm{k}}{\sqrt{14}} \end{gathered}$$