Let P be a point in the plane of the vectors A B =3 i + j - k and A...

Since P is equidistant from lines AB and AC through point A, it lies along the angle bisector direction. The unit vectors are $\hat{u}_{AB} = \frac{3\vec{i} + \vec{j} - \vec{k}}{\sqrt{11}}$ and $\hat{u}_{AC} = \frac{\vec{i} - \vec{j} + 3\vec{k}}{\sqrt{11}}$.
The bisector direction is proportional to $2\vec{i} + \vec{k}$, so $\vec{AP} = \lambda(2\vec{i} + \vec{k})$.
From $|\vec{AP}| = \frac{\sqrt{5}}{2}$: $|\lambda|\sqrt{5} = \frac{\sqrt{5}}{2}$, giving $\lambda = \frac{1}{2}$.
Thus $\vec{AP} = \vec{i} + \frac{1}{2}\vec{k}$.
Area of triangle ABP = $\frac{1}{2}|\vec{AB} \times \vec{AP}|$.
Computing: $\vec{AB} \times \vec{AP} = \frac{1}{2}\vec{i} - \frac{5}{2}\vec{j} - \vec{k}$.
$|\vec{AB} \times \vec{AP}| = \sqrt{\frac{1}{4} + \frac{25}{4} + 1} = \sqrt{\frac{30}{4}} = \frac{\sqrt{30}}{2}$.
Area = $\frac{\sqrt{30}}{4}$.