If $G$ is the centroid of a triangle having vertices $A(3h, 3k)$,$B(-3a, 0)$,and $C(3a, 0)$,then prove that $AB^2 + BC^2 + AC^2 = 3(GA^2 + GB^2 + GC^2)$.

(N/A) Let the vertices of $\Delta ABC$ be $A(3h, 3k)$,$B(-3a, 0)$,and $C(3a, 0)$.
The coordinates of the centroid $G$ are given by:
$G = \left( \frac{3h - 3a + 3a}{3}, \frac{3k + 0 + 0}{3} \right) = (h, k)$.
Now,calculate the sum of the squares of the sides:
$AB^2 = (3h - (-3a))^2 + (3k - 0)^2 = (3h + 3a)^2 + 9k^2 = 9(h+a)^2 + 9k^2 = 9(h^2 + 2ha + a^2 + k^2)$
$BC^2 = (3a - (-3a))^2 + (0 - 0)^2 = (6a)^2 = 36a^2$
$AC^2 = (3h - 3a)^2 + (3k - 0)^2 = 9(h-a)^2 + 9k^2 = 9(h^2 - 2ha + a^2 + k^2)$
Summing these:
$AB^2 + BC^2 + AC^2 = 9(h^2 + 2ha + a^2 + k^2) + 36a^2 + 9(h^2 - 2ha + a^2 + k^2)$
$= 9(2h^2 + 2a^2 + 2k^2) + 36a^2 = 18h^2 + 18k^2 + 18a^2 + 36a^2 = 18(h^2 + k^2 + 3a^2)$ ... $(1)$
Now,calculate $3(GA^2 + GB^2 + GC^2)$:
$GA^2 = (h - 3h)^2 + (k - 3k)^2 = (-2h)^2 + (-2k)^2 = 4h^2 + 4k^2$
$GB^2 = (h - (-3a))^2 + (k - 0)^2 = (h + 3a)^2 + k^2 = h^2 + 6ha + 9a^2 + k^2$
$GC^2 = (h - 3a)^2 + (k - 0)^2 = (h - 3a)^2 + k^2 = h^2 - 6ha + 9a^2 + k^2$
Summing these:
$GA^2 + GB^2 + GC^2 = (4h^2 + 4k^2) + (h^2 + 6ha + 9a^2 + k^2) + (h^2 - 6ha + 9a^2 + k^2) = 6h^2 + 6k^2 + 18a^2$
$3(GA^2 + GB^2 + GC^2) = 3(6h^2 + 6k^2 + 18a^2) = 18(h^2 + k^2 + 3a^2)$ ... $(2)$
From $(1)$ and $(2)$,we have $AB^2 + BC^2 + AC^2 = 3(GA^2 + GB^2 + GC^2)$.