Let $A(2, 2, -3)$,$B(5, 6, 9)$,and $C(2, 7, 9)$ be the vertices of a triangle. The angle bisector of $\angle A$ meets $BC$ at the point $D$. Find the coordinates of $D$.

(N/A) Let the coordinates of $D$ be $(x, y, z)$.
First,calculate the lengths of sides $AB$ and $AC$:
$AB = \sqrt{(5-2)^2 + (6-2)^2 + (9 - (-3))^2} = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$
$AC = \sqrt{(2-2)^2 + (7-2)^2 + (9 - (-3))^2} = \sqrt{0^2 + 5^2 + 12^2} = \sqrt{0 + 25 + 144} = \sqrt{169} = 13$
Since $AB = AC = 13$,$\triangle ABC$ is an isosceles triangle.
In an isosceles triangle,the angle bisector of the vertex angle is also the median to the base.
Therefore,$AD$ is the median to $BC$,which means $D$ is the midpoint of $BC$.
The coordinates of $D$ are given by the midpoint formula:
$D = \left(\frac{5+2}{2}, \frac{6+7}{2}, \frac{9+9}{2}\right) = \left(\frac{7}{2}, \frac{13}{2}, 9\right)$