The unit vector perpendicular to the vector $\hat{i}-2\hat{j}+3\hat{k}$ and coplanar with the vectors $\hat{i}+\hat{j}+\hat{k}$ and $2\hat{i}-\hat{j}-\hat{k}$ is

Let $\overrightarrow{a}=x\hat{i}+y\hat{j}+z\hat{k}$ 

$\overrightarrow{b}=\hat{i}-2\hat{j}+3\hat{k}$

$\overrightarrow{c}=\hat{i}+\hat{j}+\hat{k}$

$\overrightarrow{d}=2\hat{i}-\hat{j}-\hat{k}$

$\Rightarrow \overrightarrow{a}$is coplanar to vector $\overrightarrow{c} \& \overrightarrow{d}$  

Let $\overrightarrow{c}\times \overrightarrow{d}=\overrightarrow{E}$$$

$\overrightarrow{c}\times \overrightarrow{d}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 2& -1& -1\end{array}\right|=\hat{i}\left[-1+1\right]-\hat{j}\left[-3\right]+\hat{k}\left[-3\right]$

$\overrightarrow{E}=0\hat{i}+3\hat{j}-3\hat{k}$

Since $\overrightarrow{E}$ is perpendicular to both vectors $\overrightarrow{c}$and $\overrightarrow{d}$ so it will also be perpendicular to $\overrightarrow{a}$ as $\overrightarrow{a},\overrightarrow{c} \& \overrightarrow{d}$ are coplanar.

Now $\overrightarrow{a}\perp \overrightarrow{E} \Rightarrow \overrightarrow{a}·\overrightarrow{E}=0 \left\{a_1{a}_2+b_1{b}_2+c_1{c}_2=0\right\}$

$\left(x\hat{i}+y\hat{j}+z\hat{k}\right)·\left(0\hat{i}+3\hat{j}-3\hat{k}\right)=0$

$3y-3z=0$

$y=z ...\left(1\right)$

$\overrightarrow{a}$is also perpendicular to $\overrightarrow{b}$so, $\overrightarrow{a}·\overrightarrow{b}=0$

$\Rightarrow \left(x\hat{i}+y\hat{j}+z\hat{k}\right)·\left(\hat{i}-2\hat{j}+3\hat{k}\right)=0$

$\Rightarrow x-2y+3z=0$

$\Rightarrow x-2y+3y=0 \left\{by \left(i\right)\right\}$

$\Rightarrow x=-y ...\left(2\right)$

So $\overrightarrow{a}=\pm \left[x\hat{i}+y\hat{j}+z\hat{k}\right]$, where  $\left\{x=-y, x=-z\right\}$

$\hat{a}=\frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}=\frac{x\left[\hat{i}-\hat{j}-\hat{k}\right]}{x\sqrt{1+1+1}}=\pm \frac{1}{\sqrt{3}}\left[\hat{i}-\hat{j}-\hat{k}\right]$