If $G(2, -1, 2)$ is the centroid of tetrahedron $OABC$ where $O=(0, 0, 0)$ and $G_1$ is the centroid of $\triangle ABC$,then $\left|\overline{O G_1}\right|=$

The centroid of the triangle whose vertices are $P(1, -2, 1)$,$Q(2, 3, -1)$,and $R(1, -1, -1)$ is:

Let the centroid of a triangle formed by the points $A(4, x, 1)$,$B(y, -5, 2)$ and $C(7, 8, 3)$ be $G(3, 5, 2)$ and $CG$ meet $AB$ in $F$. Then,$F=$

In $\triangle ABC$,the midpoints of the sides $AB, BC$ and $CA$ are respectively $(l, 0, 0), (0, m, 0)$ and $(0, 0, n)$. Then,$\frac{AB^2+BC^2+CA^2}{l^2+m^2+n^2}$ is equal to

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 $A(7, -8, 1)$,$B(p, q, 5)$,and $C(q+1, 5p, 0)$ are the vertices of the triangle,then the values of $p, q, r$ are respectively . . . . . . .