The area (in square units) of the region bounded by the parabola $$y^2 = 4(x - 2)$$ and the line $$y = 2x - 8$$.

For the parabola $$y^2 = 4(x-2)$$ we write $$x = \frac{y^2}{4} + 2$$, and for the line $$y = 2x - 8$$ we have $$x = \frac{y+8}{2}$$.

Equating these expressions gives $$\frac{y^2}{4} + 2 = \frac{y+8}{2}$$, which simplifies to $$y^2 + 8 = 2y + 16 \;\Rightarrow\; y^2 - 2y - 8 = 0 \;\Rightarrow\; (y-4)(y+2) = 0$$, so $$y = 4$$ or $$y = -2$$.

The area between the curves, with the line to the right of the parabola, is $$A = \int_{-2}^{4} \left[\frac{y+8}{2} - \frac{y^2}{4} - 2\right] dy = \int_{-2}^{4} \left[\frac{y+8}{2} - \frac{y^2+8}{4}\right] dy = \int_{-2}^{4} \frac{2(y+8) - (y^2+8)}{4} dy = \int_{-2}^{4} \frac{2y + 16 - y^2 - 8}{4} dy = \int_{-2}^{4} \frac{-y^2 + 2y + 8}{4} dy = \frac{1}{4}\left[-\frac{y^3}{3} + y^2 + 8y\right]_{-2}^{4}.$$

At $$y = 4$$: $$-\frac{64}{3} + 16 + 32 = -\frac{64}{3} + 48 = \frac{-64 + 144}{3} = \frac{80}{3}$$, and at $$y = -2$$: $$\frac{8}{3} + 4 - 16 = \frac{8}{3} - 12 = \frac{8 - 36}{3} = -\frac{28}{3}$$. Hence, $$A = \frac{1}{4}\left[\frac{80}{3} - \left(-\frac{28}{3}\right)\right] = \frac{1}{4} \cdot \frac{108}{3} = \frac{108}{12} = 9.$$ The answer is Option (2): $$\boxed{9}$$.