The extremities of a diagonal of a parallelogram are the points $(3, -4)$ and $(-6, 5)$. If the third vertex is $(-2, 1)$,then the fourth vertex is:

The midpoints of the sides of a triangle are $(1, 5, -1)$,$(0, 4, -2)$,and $(2, 3, 4)$. Find the centroid of the triangle.

$A$ 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

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 $D$.

The centroid of a tetrahedron with vertices $A(3, -5, x)$,$B(5, 4, 2)$,$C(7, -7, y)$,and $D(1, 0, z)$ is $G(4, -2, 2)$. Then,the value of $x + y + z$ is:

$A(2,3,-4), B(-3,3,-2), C(-1,4,2), D(3,5,1)$ are the vertices of a tetrahedron. If $G_1, G_2$ and $G_3$ are the centroids of the three faces having the vertex $D$ in common,then the centroid of the $\Delta G_1 G_2 G_3$ is