Let the maximum and minimum values of $$\left(\sqrt{8x - x^2 - 12} - 4\right)^2 + (x - 7)^2$$, $$x \in \mathbb{R}$$ be $$M$$ and $$m$$, respectively. Then $$M^2 - m^2$$ is equal to _________

Given expression: $$D^2 = (x - 7)^2 + (y - 4)^2$$

Where the point $$(x, y)$$ lies on the curve:

$$y = \sqrt{8x - x^2 - 12} \implies y^2 = -(x^2 - 8x + 12)$$

$$(x - 4)^2 + y^2 = 4 \quad (y \ge 0)$$

This represents a semi-circle with center $$C(4, 0)$$ and radius $$R = 2$$.

Distance from center $$C(4,0)$$ to the point $$P(7,4)$$:

$$PC = \sqrt{(7-4)^2 + (4-0)^2} = 5$$

The nearest point on the semi-circle is $$A$$, lying on the line segment $$CP$$:

$$d_{\min} = PA = PC - R = 5 - 2 = 3 \implies m = d_{\min}^2 = 9$$

The farthest point on the semi-circle is the left boundary endpoint $$B(2, 0)$$:

$$d_{\max}^2 = PB^2 = (2 - 7)^2 + (0 - 4)^2 = 25 + 16 = 41 \implies M = d_{\max}^2 = 41$$

$$M^2 - m^2 = 41^2 - 9^2 = 1681 - 81 = 1600$$