The position vectors of the vertices A, B and C of a triangle are $$2\hat{i} - 3\hat{j} + 3\hat{k}$$, $$2\hat{i} + 2\hat{j} + 3\hat{k}$$ and $$-\hat{i} + \hat{j} + 3\hat{k}$$ respectively. Let $$l$$ denotes the length of the angle bisector AD of $$\angle BAC$$ where D is on the line segment BC, then $$2l^2$$ equals :

$$A(2,-3,3)$$, $$B(2,2,3)$$, $$C(-1,1,3)$$.

$$\vec{AB} = (0,5,0)$$, $$AB = 5$$.

$$\vec{AC} = (-3,4,0)$$, $$AC = 5$$.

Since $$AB = AC$$, D is the midpoint of BC (angle bisector theorem with equal sides).

$$D = \frac{B + C}{2} = \left(\frac{1}{2}, \frac{3}{2}, 3\right)$$

$$\vec{AD} = \left(\frac{1}{2} - 2, \frac{3}{2}+3, 0\right) = \left(-\frac{3}{2}, \frac{9}{2}, 0\right)$$

$$l^2 = \frac{9}{4} + \frac{81}{4} = \frac{90}{4} = \frac{45}{2}$$

$$2l^2 = 45$$.

The answer is $$45$$, which corresponds to Option (4).