The unit vector which is orthogonal to the vector $3\overline{\mathrm{i}}+2\overline{\mathrm{j}}+6\overline{\mathrm{k}}$ and is co planar with the vectors $2\overline{\mathrm{i}}+\overline{\mathrm{j}}+\overline{\mathrm{k}} \mathrm{and} \overline{\mathrm{i}}-\overline{\mathrm{j}}+\overline{\mathrm{k}} \mathrm{is}$

$\begin{array}{rcl}\overline{\mathrm{a}}& =& 2\mathrm{i}+\mathrm{j}+\mathrm{k} , \overline{\mathrm{b}}=\mathrm{i}-\mathrm{j}+\mathrm{k}\\ \overline{\mathrm{c}}& =& 3\mathrm{i}+2\mathrm{j}+6\mathrm{k} , \overline{\mathrm{d}}=\mathrm{xi}+\mathrm{yj}+\mathrm{zk}\\ [\overline{\mathrm{d}} \overline{\mathrm{a}} \overline{\mathrm{b}}]& =& 0 \Rightarrow \left|\begin{array}{ccc}\mathrm{x}& \mathrm{y}& \mathrm{z}\\ 2& 1& 1\\ 1& -1& 1\end{array}\right|=0\\ & \Rightarrow & \mathrm{x}(1+1)-\mathrm{y}(2-1)+\mathrm{z}(-2-1)=0\\ 2\mathrm{x}-\mathrm{y}-3\mathrm{z}& =& 0 ···\left(1\right)\\ \overline{\mathrm{d}}·\overline{\mathrm{c}}& =& 0 \Rightarrow 3\mathrm{x}+2\mathrm{y}+6\mathrm{z}=0 ···\left(2\right)\end{array}$

$$\begin{gathered} \left(2\right) + 2\times \left(1\right) \Rightarrow 3\mathrm{x}+2\mathrm{y}+6\mathrm{z}=0 \\ \underline{ 4\mathrm{x}-2\mathrm{y}-6\mathrm{z}=0} \\ 7\mathrm{x}=0 \Rightarrow \mathrm{x}=0 \\ \Rightarrow \mathrm{y}=-3\mathrm{z} \\ \left|\overline{\mathrm{d}}\right| =1 \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2=1 \\ {\left(-3z\right)}^2+{\mathrm{z}}^2=1 \Rightarrow \mathrm{z}=\pm \frac{1}{\sqrt{10}} \\ \overline{\mathrm{d}} =\pm \frac{1}{\sqrt{10}} (3\overline{j}-\overline{k}) \end{gathered}$$