The points $(-2, 4, 7)$,$(3, -6, -8)$,and $(1, -2, -2)$ are:

If $G(3, -5, r)$ is the centroid of $\triangle ABC$,where $A \equiv (7, -8, 1)$,$B \equiv (p, q, 5)$,and $C \equiv (q+1, 5p, 0)$ are vertices of the triangle $ABC$,then the values of $p, q, r$ are respectively:

Let $ABCD$ be a parallelogram and $E$ be the mid-point of $AB$. If $P$ is the point of intersection of $DE$ and $AC$,then $\frac{DP}{PE} + \frac{AP}{PC} = $

Let $A(4, -2)$,$B(1, 1)$,and $C(9, -3)$ be the vertices of a triangle $ABC$. Then the maximum area of the parallelogram $AFDE$,formed with vertices $D, E$,and $F$ on the sides $BC, CA$,and $AB$ of the triangle $ABC$ respectively,is $\qquad$ .

The mid-points of the sides of a triangle are $(1, 5, -1), (0, 4, -2)$ and $(2, 3, 4)$. Find its vertices. Also,find the centroid of the triangle.

The points $(2, 3, 4)$,$(-1, -2, 1)$,and $(5, 8, 7)$ are