$G(1,0,1)$ is the centroid of the triangle $ABC$. If $A=(1,-4,2)$ and $B=(3,1,0)$,then $AG^2+CG^2=$

$A$ point moves in such a way that the sum of its distances from the $xy$-plane and $yz$-plane remains equal to its distance from the $zx$-plane. The locus of the point is:

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:

Three consecutive vertices of a parallelogram $ABCD$ are $A(6, -2, 4)$,$B(2, 4, -8)$,and $C(-2, 2, 4)$. Find the coordinates of the fourth vertex $D$.

Verify that the points $A(-1, 2, 1)$,$B(1, -2, 5)$,$C(4, -7, 8)$,and $D(2, -3, 4)$ are the vertices of a parallelogram.

If $A \equiv (5, 1, p)$,$B \equiv (1, q, p)$,and $C \equiv (1, -2, 3)$ are the vertices of a triangle and $G \equiv (r, -\frac{4}{3}, \frac{1}{3})$ is its centroid,then the values of $p, q, r$ are respectively: