The points $A(-1, 2, 3)$,$B(2, -3, 1)$,and $C(3, 1, -2)$:

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 $D$.

If $G(4, 3, 3)$ is the centroid of the triangle $ABC$ whose vertices are $A(a, 3, 1)$,$B(4, 5, b)$,and $C(6, c, 5)$,then the values of $a, b, c$ are:

If the coordinates of the points $A, B, C$ are $(-1, 3, 2)$,$(2, 3, 5)$,and $(3, 5, -2)$ respectively,then $\angle A = \dots^o$.

If the points $P = \hat{i} + 2 \hat{j}$,$Q = 4 \hat{i} + 6 \hat{j}$,$R = 5 \hat{i} + 7 \hat{j}$,and $S = a \hat{i} + b \hat{j}$ are the consecutive vertices of a parallelogram $PQRS$,then:

$A$ parallelogram has vertices $A(4,4,-1)$,$B(5,6,-1)$,$C(6,5,1)$ and $D(x, y, z)$. Then the vertex $D$ is