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

If the vertices of a triangle $ABC$ are $A(1, 2, 3)$,$B(h, -3, 0)$,and $C(-4, k, -1)$ and the centroid of the triangle is $\left(5, -1, \frac{2}{3}\right)$,then triangle $ABC$ is

If the points $A(0, 1, 2)$,$B(2, -1, 3)$,and $C(1, -3, 1)$ are the vertices of a triangle,then the triangle is

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:

If $a>0, b>0$, then the maximum area of the parallelogram whose three vertices are $O(0,0)$, $A(a \cos \theta, b \sin \theta)$, and $B(a \cos \theta, -b \sin \theta)$ is

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