The centroid of a triangle with vertices $A(3,4,5)$, $B(6,7,2)$, and $C(x, y, z)$ is $(3,2,3)$. Then $x+y+z=$

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

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 are $(1, 3)$,$(2, 0)$,and $(5, 1)$. Find its fourth vertex.

Verify that the points $(0, 7, -10)$,$(1, 6, -6)$,and $(4, 9, -6)$ are the vertices of an isosceles triangle.

If the origin is the centroid of a triangle $ABC$ having vertices $A(a, 1, 3)$,$B(-2, b, -5)$,and $C(4, 7, c)$,find the values of $a, b, c$.