A unit vector perpendicular to $\overrightarrow{A}=2\hat{i}+3\hat{j}+\hat{k}$  and $\overrightarrow{B}=\hat{i}-\hat{j}+\hat{k}$  both is:

Let $\overrightarrow{R}$  is the vector perpendicular to $\overrightarrow{A}$  and $\overrightarrow{B}$  both such

That $\overrightarrow{R}=\overrightarrow{A}\times \overrightarrow{B}$

$\hat{R}=\frac{\overrightarrow{R}}{R}=\frac{\overrightarrow{A}\times \overrightarrow{B}}{\left|\overrightarrow{A}\times \overrightarrow{B}\right|}$

$$\begin{gathered} \overrightarrow{A}\times \overrightarrow{B}=\left|\stackrel{⏜}{i } \stackrel{⏜}{j } \stackrel{⏜}{k } \\ 2 3 1 \\ 1 -1 1\right| \end{gathered}$$

$=\hat{i}\left(3+1\right)-\hat{j}\left(2-1\right)+\hat{k}\left(-2-3\right)$

$=4\hat{i}-\hat{j}-5\hat{k}$

And $\left|\overrightarrow{A}\times \overrightarrow{B}\right|=\sqrt{{\left(4\right)}^2+{\left(-1\right)}^2+{\left(-5\right)}^2}=\sqrt{42}$

Thus, $\hat{R}=\frac{4\hat{i}-\hat{j}-5\hat{k}}{\sqrt{42}}$