If the points $(x, -1)$,$(3, y)$,$(-2, 3)$,and $(-3, -2)$ are the vertices of a parallelogram,then:

$A(2,3,5), B(\alpha, 3,3)$ and $C(7,5, \beta)$ are the vertices of a triangle. If the median through $A$ is equally inclined with the coordinate axes,then $\frac{\beta}{\alpha}=$

$A(2, 3, k)$, $B(-1, k, -1)$, and $C(4, -3, 2)$ are the vertices of $\triangle ABC$. If $AB = AC$ and $k > 0$, then $\triangle ABC$ is:

The centroid of a triangle $ABC$ is at the point $(1,1,1)$. If the coordinates of $A$ and $B$ are $(3,-5,7)$ and $(-1,7,-6)$ respectively,find the coordinates of the point $C$.

If the origin is the centroid of the triangle whose vertices are $A(2, p, -3)$,$B(q, -2, 5)$,and $C(-5, 1, r)$,then

The incenter and centroid of the triangle,whose vertices are $A \equiv(0,3,0), B \equiv(0,0,4)$,and $C \equiv(0,3,4)$,are respectively given by