1. Write the coordinates of the points

Given position vectors: $A=(2,-3,3),\quad B=(2,2,3),\quad C=(-1,1,3)$

All three points have $z=3$, so the triangle lies in a plane parallel to the $xy$-plane.

2. Find the side lengths adjacent to angle $A$

We need the lengths of $AB$ and $AC$.

$\overrightarrow{AB}=B-A=(2-2,\,2-(-3),\,3-3)=(0,5,0)$ $|AB|=5$

$\overrightarrow{AC}=C-A=(-1-2,\,1-(-3),\,3-3)=(-3,4,0)$ $|AC|=\sqrt{(-3)^2+4^2}=5$

So, $AB=AC=5$

Thus triangle $ABC$ is isosceles at $A$.

3. Use the property of angle bisector in an isosceles triangle

Since $AB=AC$, the angle bisector from $A$ to side $BC$ also bisects $BC$. Hence $D$ is the midpoint of $BC$.

Midpoint of $B(2,2,3)$ and $C(-1,1,3)$ is $D=\left(\frac{2+(-1)}{2},\frac{2+1}{2},\frac{3+3}{2}\right)=\left(\frac12,\frac32,3\right)$

4. Find the length of the angle bisector $AD$

$\overrightarrow{AD}=D-A=\left(\frac12-2,\frac32-(-3),3-3\right)=\left(-\frac32,\frac92,0\right)$

Therefore, $l^2=|AD|^2=\left(-\frac32\right)^2+\left(\frac92\right)^2=\frac{9}{4}+\frac{81}{4}=\frac{90}{4}=\frac{45}{2}$

So, $2l^2=2\cdot \frac{45}{2}=45$

5. Check options

$2l^2=45$ So the correct option is:

A: 45

6. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.