If the points $(-1, -1, 2), (2, m, 5)$ and $(3, 11, 6)$ are collinear,find the value of $m$.

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:

The centroid of a tetrahedron with vertices $A(3, -5, x)$,$B(5, 4, 2)$,$C(7, -7, y)$,and $D(1, 0, z)$ is $G(4, -2, 2)$. Then,the value of $x + y + z$ is:

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 $\frac{AB^2+BC^2+CA^2}{l^2+m^2+n^2}=$

The incenter and centroid of the triangle,whose vertices are $A \equiv(0,3,0), B \equiv(0,0,4)$,and $C \equiv(0,3,4)$,are respectively given by

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