Let the three given points be A(3, 0), B(-2, -2) and C(8, 2).

Finding the equation of line through A and B:

Using the equation of line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$:

$$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$$

Substituting A(3, 0) and B(-2, -2):

$$y - 0 = \frac{-2 - 0}{-2 - 3}(x - 3)$$ $$y = \frac{-2}{-5}(x - 3)$$ $$y = \frac{2}{5}(x - 3)$$

Checking if point C lies on this line:

Substituting $x = 8$ in the equation:

$$y = \frac{2}{5}(8 - 3) = \frac{2}{5} \times 5 = 2$$

Since the $y$-coordinate of point C is 2, and substituting $x = 8$ gives $y = 2$, point C satisfies the equation of line AB.

Conclusion:

The point $(x, y)$ lies on the line with slope $m$ and through the fixed point $(x_0, y_0)$, if and only if its coordinates satisfy the equation. Since point C(8, 2) satisfies the equation of line passing through A and B, all three points lie on the same line. Hence, the three points $(3, 0), (-2, -2)$ and $(8, 2)$ are collinear.