To show that the points $(3,0), (-2,-2),$ and $(8,2)$ are collinear,it suffices to show that the line passing through points $(3,0)$ and $(-2,-2)$ also passes through the point $(8,2)$.
The equation of the line passing through points $(3,0)$ and $(-2,-2)$ is given by the two-point form:
$(y - 0) = \frac{-2 - 0}{-2 - 3}(x - 3)$
$y = \frac{-2}{-5}(x - 3)$
$y = \frac{2}{5}(x - 3)$
$5y = 2x - 6$
$2x - 5y = 6$
Now,we check if the point $(8,2)$ satisfies this equation:
$L.H.S. = 2(8) - 5(2) = 16 - 10 = 6$
Since $L.H.S. = R.H.S. = 6$,the point $(8,2)$ lies on the line passing through $(3,0)$ and $(-2,-2)$.
Therefore,the points $(3,0), (-2,-2),$ and $(8,2)$ are collinear.