If the function $$f(x) = \begin{cases} \frac{72^x - 9^x - 8^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}, & x \neq 0 \\ a \log_e 2 \log_e 3, & x = 0 \end{cases}$$ is continuous at $$x = 0$$, then the value of $$a^2$$ is equal to

$$\frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}$$.

Numerator: $$72^x - 9^x - 8^x + 1 = (9^x-1)(8^x-1) \approx (x\ln 9)(x\ln 8) = x^2 \ln 9 \cdot \ln 8$$ near $$x = 0$$.

Denominator: $$\sqrt{2}-\sqrt{1+\cos x} = \sqrt{2} - \sqrt{2}\cos(x/2)$$ (using $$1+\cos x = 2\cos^2(x/2)$$).

$$= \sqrt{2}(1 - \cos(x/2)) \approx \sqrt{2} \cdot \frac{x^2}{8}$$

Limit: $$\frac{x^2 \ln 9 \cdot \ln 8}{\sqrt{2}x^2/8} = \frac{8 \cdot 2\ln 3 \cdot 3\ln 2}{\sqrt{2}} = \frac{48\ln 2 \cdot \ln 3}{\sqrt{2}}$$.

For continuity: $$a \cdot \ln 2 \cdot \ln 3 = \frac{48\ln 2 \cdot \ln 3}{\sqrt{2}}$$, so $$a = \frac{48}{\sqrt{2}} = 24\sqrt{2}$$.

$$a^2 = 576 \times 2 = 1152$$.

The correct answer is Option 2: 1152.