Find the coordinates of the points which trisect the line segment joining the points $A(2, 1, -3)$ and $B(5, -8, 3)$.

(N/A) Let $P(x_{1}, y_{1}, z_{1})$ and $Q(x_{2}, y_{2}, z_{2})$ be the points that trisect the line segment $AB$.
Since point $P$ divides $AB$ in the ratio $1:2$ internally,we have:
$P = \left(\frac{1(5) + 2(2)}{1 + 2}, \frac{1(-8) + 2(1)}{1 + 2}, \frac{1(3) + 2(-3)}{1 + 2}\right)$
$P = \left(\frac{5 + 4}{3}, \frac{-8 + 2}{3}, \frac{3 - 6}{3}\right)$
$P = \left(\frac{9}{3}, \frac{-6}{3}, \frac{-3}{3}\right) = (3, -2, -1)$
Since point $Q$ divides $AB$ in the ratio $2:1$ internally,we have:
$Q = \left(\frac{2(5) + 1(2)}{2 + 1}, \frac{2(-8) + 1(1)}{2 + 1}, \frac{2(3) + 1(-3)}{2 + 1}\right)$
$Q = \left(\frac{10 + 2}{3}, \frac{-16 + 1}{3}, \frac{6 - 3}{3}\right)$
$Q = \left(\frac{12}{3}, \frac{-15}{3}, \frac{3}{3}\right) = (4, -5, 1)$
Thus,the coordinates of the points are $(3, -2, -1)$ and $(4, -5, 1)$.