A unit vector perpendicular to the plane of $\vec a=2\hat i-6\hat j-3\hat k$ and $\vec b = 4\hat i +3\hat j-\hat k$, is

$\frac{1}{7}(3\hat i +2\hat j+6\hat k)$

We have,

$\vec a×\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-6&-3\\4&3&-1\end{vmatrix}=5(3\hat i-2\hat j+6\hat k)$

$∴|\vec a×\vec b|=5\sqrt{9+4+36}=35$

Hence, required unit vector $\hat n$ is given by

$\hat n=\frac{4}{35}(3\hat i-2\hat j+6\hat k)=\frac{1}{7}(3\hat i +2\hat j+6\hat k)$