Let $\vec{OA} =\hat{ i} +3\hat{ j}-2 \hat{ k}$ and $\vec{OB}=3\hat {i}+\hat {j}-2\hat {k}$. The vector $\vec{OC}$ bisecting the angle AOB and C being a point on the line AB is

$2 (\hat i+\hat j-\hat k)$

Taking O as the origin, the position vectors of A and B are $\vec a =\hat i +3\hat j-2 \hat k$ and $\vec b = 3\hat i+\hat j-2\hat k$ respectively.

We have, $|\vec a|=|\vec b|=\sqrt{14}$.

So, the bisector OC of ∠AOB meets AB at its mid-point C.

$∴\vec{OC}=\frac{1}{2}(\vec{OA}+\vec{OB})=2 (\hat i+\hat j-\hat k)$