Let $$S = \{x \in [-\frac{\pi}{2}, \frac{\pi}{2}]: 9^{1-\tan^2 x} + 9^{\tan^2 x} = 10\}$$ and $$\beta = \sum_{x \in S} \tan^2 \frac{x}{3}$$, then $$\frac{1}{6}(\beta - 14)^2$$ is equal to

By setting $$t = 9^{\tan^2 x}$$ it follows that $$9^{1-\tan^2 x} = \frac{9}{t}$$. Hence the equation becomes $$\frac{9}{t} + t = 10 \implies t^2 - 10t + 9 = 0 \implies (t-1)(t-9) = 0$$ so that $$t=1$$ or $$t=9$$. If $$t=1$$ then $$\tan^2 x=0$$ giving $$x=0$$, while if $$t=9$$ then $$\tan^2 x=1$$ which leads to $$x=\pm\frac{\pi}{4}$$. Therefore the set of solutions is $$S = \left\{-\frac{\pi}{4}, 0, \frac{\pi}{4}\right\}$$.

The value of $$\beta$$ can be written as $$\beta = \tan^2\left(-\frac{\pi}{12}\right) + \tan^2(0) + \tan^2\left(\frac{\pi}{12}\right) = 2\tan^2\frac{\pi}{12}$$. Since $$\tan 15° = 2 - \sqrt{3}$$, it follows that $$\tan^2 15° = 7 - 4\sqrt{3}$$ and therefore $$\beta = 14 - 8\sqrt{3}$$.

Finally, evaluating the desired expression gives
$$\frac{1}{6}(\beta - 14)^2 = \frac{1}{6}(-8\sqrt{3})^2 = \frac{192}{6} = \textbf{32}$$