$A$ parallelogram has vertices $A(4,4,-1)$,$B(5,6,-1)$,$C(6,5,1)$ and $D(x, y, z)$. Then the vertex $D$ 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,$F=$

Consider a cuboid where all edges are integers and the base is a square. Suppose the sum of all its edges is numerically equal to the sum of the areas of all its six faces. Then,the sum of all its edges is

$A(2,3,5), B(\alpha, 3,3)$ and $C(7,5, \beta)$ are the vertices of a triangle. If the median through $A$ is equally inclined with the coordinate axes,then $\frac{\beta}{\alpha}=$

In $\triangle ABC$,if the midpoints of the sides $AB, BC$ and $CA$ are respectively $(l, 0, 0), (0, m, 0)$ and $(0, 0, n)$,then $\frac{AB^2+BC^2+CA^2}{l^2+m^2+n^2}=$

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: