If $\overrightarrow{\mathrm{a}}=\mathrm{\lambda }(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}),\overrightarrow{\mathrm{b}}=\mathrm{\mu }(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})$, and $\overrightarrow{c}$ are unit vectors . $\overrightarrow{c}$perpendicular to the vector $\overrightarrow{a}$ and coplanar with $\overrightarrow{a}$ and $\overrightarrow{b}$, then a  unit vector $\overrightarrow{d}$ perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{c}$ is

Let $\overrightarrow{\mathrm{c}}=x\hat{i}+y\hat{j}+z\hat{k}$. since a,b,c are coplanar , so
$\left|\begin{array}{ccr}\mathrm{x}& \mathrm{y}& \mathrm{z}\\ 1& 1& -1\\ 1& -1& 1\end{array}\right|=0\Rightarrow 0\cdot \mathrm{x}-2\mathrm{y}-2\mathrm{z}=0\Rightarrow \mathrm{y}+\mathrm{z}=0$.
Also $\overrightarrow{c}$ is perpendicular to $\overrightarrow{a}$, so $\mathrm{x}+\mathrm{y}-\mathrm{z}=0$. Therefore $\frac{\mathrm{x}}{-2}=\frac{\mathrm{y}}{1}=\frac{\mathrm{z}}{-1}$ and ${\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2=1$ ( $\overrightarrow{c}$ being a unit vector). We have $\mathrm{x}=-\frac{2}{\sqrt{6}},\mathrm{y}=\frac{1}{\sqrt{6}},\mathrm{z}=-\frac{1}{\sqrt{6}}$.
Thus $$\begin{gathered} \overrightarrow{\mathrm{c}}=\frac{1}{\sqrt{6}}(-2\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ \overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{c}}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& -1\\ -2& 1& -1\end{array}\right|=(0\mathrm{i}+3\mathrm{j}+3\mathrm{k}) \end{gathered}$$
$$\begin{gathered} \overrightarrow{\mathrm{d}}=\frac{1}{|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{c}}|}\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{c}} \\ \overrightarrow{\mathrm{d}}=\frac{(0\mathrm{i}+3\mathrm{j}+3\mathrm{k})}{\sqrt{9+9}}=\frac{1}{\sqrt{2}}(\mathrm{j}+\mathrm{k})\text{. } \end{gathered}$$