If $\overrightarrow{a}=\overrightarrow{i}-\overrightarrow{2j}+\overrightarrow{3k}$ and $\overrightarrow{b}=\overrightarrow{3i}+\overrightarrow{j}+\overrightarrow{2k}$ then a unit vector perpendicular to $\overrightarrow{a}$ and $\overrightarrow{b}$ is

$\begin{array}{1 1}(1)\large\frac{\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}}{\sqrt{3}}&(2)\large\frac{\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k}}{\sqrt{3}}\\(3)\large\frac{-\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{2k}}{\sqrt{3}}&(4)\large\frac{\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k}}{\sqrt{3}}\end{array}$

$\overrightarrow{a}=\overrightarrow{i}-\overrightarrow{2j}+\overrightarrow{3k}$ and $\overrightarrow{b}=\overrightarrow{3i}+\overrightarrow{j}+\overrightarrow{2k}$
$\overrightarrow{a} \times \overrightarrow{b}= \begin {vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ 1 & -2 & 3 \\ 3 & 1 & 2 \end {vmatrix}$
$\qquad = \overrightarrow{i}(-4-3) - \overrightarrow{j} (2-9) +\overrightarrow{k} (1+6)$
$\qquad= -7\overrightarrow{i} +7\overrightarrow{j} +7\overrightarrow{k}$
$\overrightarrow{a} \times \overrightarrow{b}= \sqrt {(-7)^2 +7^2+7^2} =7 \sqrt {3}$
The unit vector perpendicular to $\overrightarrow{a}$ & $\overrightarrow{b}$ is
$\qquad = \pm \large\frac{\overrightarrow{a} \times \overrightarrow{b}}{|\overrightarrow{a} \times \overrightarrow{b}|}$
$\qquad=\pm \large\frac{7 (-\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k})}{7 \sqrt {3}}$
$\qquad=\pm \large\frac{ (-\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k})}{\sqrt {3}}$
$\qquad= \large\frac{\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k}}{\sqrt{3}}$
Hence 4 is the correct answer