The area bounded by the curve $$y = |x^2 - 9|$$ and the line $$y = 3$$ is

We need to find the area bounded by $$y = |x^2 - 9|$$ and $$y = 3$$.

First, we find the intersection points. In the first case, $$x^2 - 9 = 3 \Rightarrow x^2 = 12 \Rightarrow x = \pm 2\sqrt{3}$$, and in the second case, $$-(x^2 - 9) = 3 \Rightarrow x^2 = 6 \Rightarrow x = \pm \sqrt{6}$$.

Since $$|x^2 - 9| = \begin{cases} 9 - x^2 & |x| \leq 3 \\ x^2 - 9 & |x| > 3 \end{cases},$$ for $$|x| \leq 3$$ we have $$|x^2 - 9| = 9 - x^2$$ which is above $$y = 3$$ when $$9 - x^2 \geq 3$$, i.e., $$|x| \leq \sqrt{6}$$. For $$|x| > 3$$ the expression becomes $$x^2 - 9$$, which equals 3 at $$|x| = 2\sqrt{3}$$.

By symmetry, the required area is $$2\Bigl[\int_0^{\sqrt{6}} \bigl((9 - x^2) - 3\bigr)\,dx + \int_{\sqrt{6}}^{3} \bigl(3 - (9 - x^2)\bigr)\,dx + \int_3^{2\sqrt{3}} \bigl(3 - (x^2 - 9)\bigr)\,dx\Bigr]$$ which simplifies to $$2\Bigl[\int_0^{\sqrt{6}} (6 - x^2)\,dx + \int_{\sqrt{6}}^{3} (x^2 - 6)\,dx + \int_3^{2\sqrt{3}} (12 - x^2)\,dx\Bigr].$$

Evaluating each integral, we find $$I_1 = \int_0^{\sqrt{6}} (6 - x^2)\,dx = \left[6x - \frac{x^3}{3}\right]_0^{\sqrt{6}} = 6\sqrt{6} - 2\sqrt{6} = 4\sqrt{6},$$ $$I_2 = \int_{\sqrt{6}}^{3} (x^2 - 6)\,dx = \left[\frac{x^3}{3} - 6x\right]_{\sqrt{6}}^{3} = (9 - 18) - (2\sqrt{6} - 6\sqrt{6}) = -9 + 4\sqrt{6},$$ $$I_3 = \int_3^{2\sqrt{3}} (12 - x^2)\,dx = \left[12x - \frac{x^3}{3}\right]_3^{2\sqrt{3}} = \bigl(24\sqrt{3} - 8\sqrt{3}\bigr) - (36 - 9) = 16\sqrt{3} - 27.$$

Therefore, the total area is $$\text{Area} = 2\bigl(4\sqrt{6} + (-9 + 4\sqrt{6}) + (16\sqrt{3} - 27)\bigr) = 2(8\sqrt{6} + 16\sqrt{3} - 36) = 16\sqrt{6} + 32\sqrt{3} - 72.$$

Note: $$32\sqrt{3} = 16\sqrt{3}\cdot 2 = 16\sqrt{12}/\sqrt{1}$$. Actually, $$\sqrt{12} = 2\sqrt{3}$$, so $$16\sqrt{12} = 32\sqrt{3}$$.

Hence, the required area is $$16\sqrt{6} + 16\sqrt{12} - 72$$, which corresponds to Option C.