If $\overline{\mathrm{a}}=\overline{\mathrm{i}}+\overline{\mathrm{j}}-\overline{\mathrm{k}}, \overline{\mathrm{b}}=\overline{\mathrm{i}}-\overline{\mathrm{j}}+\overline{\mathrm{k}}, \overline{\mathrm{c}}$ is a unit vector perpendicular to $\overline{\mathrm{a}}$ and coplanar with $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ ,then $\overline{\mathrm{c}}=$

$\begin{array}{rcl}\overline{\mathrm{c}}& =& \mathrm{xi}+\mathrm{yj}+\mathrm{zk}\end{array}$

$\begin{array}{rcl}\overline{\mathrm{c}}·\overline{\mathrm{a}}& =& 0 \Rightarrow \mathrm{x}+\mathrm{y}-\mathrm{z}=0 -\left(1\right)\end{array}$

$\begin{array}{rcl}[\overline{\mathrm{c}} \overline{\mathrm{a}} \overline{\mathrm{b}}]& =& 0 \Rightarrow \left|\begin{array}{ccc}\mathrm{x}& \mathrm{y}& \mathrm{z}\\ 1& 1& -1\\ 1& -1& 1\end{array}\right|=0\end{array}$

$\begin{array}{rcl}\mathrm{x}(1-1)-\mathrm{y}(1+1)+\mathrm{z}(-1-1)& =& 0 \Rightarrow \mathrm{y}=\mathrm{z}\end{array}$

$\begin{array}{rcl}\mathrm{x}+\mathrm{y}+\mathrm{y}& =& 0 \Rightarrow \mathrm{x}=-2\mathrm{y}\\ \left|\overline{\mathrm{c}}\right|& =& 1 \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2=1 \Rightarrow 4{\mathrm{y}}^2+{\mathrm{y}}^2+{\mathrm{y}}^2=1\\ & \Rightarrow & \mathrm{y}=\pm \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{6}$}\right.\\ \overline{\mathrm{c}}& =& \pm \frac{1}{\sqrt{6}} (-2\mathrm{i}+\mathrm{j}-\mathrm{k})\end{array}$