If the centroid of $\Delta ABC$ is $(0,0,0)$,where $A(1,1,1), B(2,1,2), C(x, y, z)$,then $(x, y, z) = \ldots \ldots$

If the points $(2,4,-1), (3,6,-1)$ and $(4,5,1)$ are three consecutive vertices of a parallelogram,then its fourth vertex is

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

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 . . . . . . .

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

Three vertices of a parallelogram $ABCD$ are $A(3, -1, 2)$,$B(1, 2, -4)$,and $C(-1, 1, 2)$. Find the coordinates of the fourth vertex $D$.