Question - Finding the General Coordinates of a Point in a Triangle

Let the coordinates of A be $(x_1, y_1, z_1)$ and B be $(x_2, y_2, z_2)$, and let C be $(x_3, y_3, z_3)$. The centroid G of triangle ABC has coordinates given by:

$G \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3} \right)$

Given that G is at (1, 1, 1), we have:

$1 = \frac{x_1+x_2+x_3}{3}$

$1 = \frac{y_1+y_2+y_3}{3}$

$1 = \frac{z_1+z_2+z_3}{3}$

Given the coordinates of A $(x_1, y_1, z_1) = (3, -5, 7)$ and B $(x_2, y_2, z_2) = (-1, 7, 6)$, we can substitute them into the equations:

$1 = \frac{3 - 1 + x_3}{3}$

$1 = \frac{-5 + 7 + y_3}{3}$

$1 = \frac{7 + 6 + z_3}{3}$

Now solve for $x_3, y_3, z_3$:

$1 = \frac{2 + x_3}{3}$ -> Multiply both sides by 3 -> $3 = 2 + x_3$ -> $x_3 = 1$

$1 = \frac{2 + y_3}{3}$ -> Multiply both sides by 3 -> $3 = 2 + y_3$ -> $y_3 = 1$

$1 = \frac{13 + z_3}{3}$ -> Multiply both sides by 3 -> $3 = 13 + z_3$ -> $z_3 = -10$

Thus, the coordinates of point C are $(x_3, y_3, z_3) = (1, 1, -10)$.