Let $\overline{\mathrm{a}} = 2\overline{\mathrm{i}} + \overline{\mathrm{j}} + \overline{\mathrm{k}} , \overline{\mathrm{b}} = \overline{\mathrm{i}}+ 2\overline{ \mathrm{j}}-\overline{\mathrm{k}}$ and a unit vector $\overline{\mathrm{c}}$ be co planar. If $\overline{\mathrm{c}}$ is perpendicular to $\overline{\mathrm{a}}$ then

 $\overline{\mathrm{c}}$ =

$\begin{array}{rcl}\overline{\mathrm{c}}& =& \mathrm{xi}+\mathrm{yj}+\mathrm{zk}\\ [\overline{\mathrm{c}} \overline{\mathrm{a}} \overline{\mathrm{b}}]& =& 0 \Rightarrow \left|\begin{array}{ccc}\mathrm{x}& \mathrm{y}& \mathrm{z}\\ 2& 1& 1\\ 1& 2& -1\end{array}\right|=0\end{array}$

$\begin{array}{rcl}\mathrm{x}(-1-2)-\mathrm{y}(-2-1)+\mathrm{z}(4-1)& =& 0\end{array}$

$\begin{array}{rcl}-3\mathrm{x}+3\mathrm{y}+3\mathrm{z}& =& 0 \Rightarrow -\mathrm{x}+\mathrm{y}+\mathrm{z}=0 -\left(1\right)\\ \overline{\mathrm{c}}·\overline{\mathrm{a}}& =& 0 \Rightarrow 2\mathrm{x}+\mathrm{y}+\mathrm{z}=0 -\left(2\right)\\ \left(2\right) - \left(1\right) & \Rightarrow & 3\mathrm{x}=0\Rightarrow \mathrm{x}=0, \mathrm{y}=-\mathrm{z}\\ \left|\overline{\mathrm{c}}\right|& =& 1 \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2=1 \Rightarrow 2{\mathrm{y}}^2=1 \Rightarrow \mathrm{y}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.\\ \overline{\mathrm{c}}& =& \pm \frac{1}{\sqrt{2}} (\mathrm{j}-\mathrm{k})\end{array}$