We have, $S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^{2} x}+9^{\tan ^{2} x}=10\right\}$ and \beta=\sum_\limits{x \in S} \tan ^{2}\left(\frac{x}{3}\right) $\begin{aligned} & 9^{1-\tan ^2 x}+9^{\tan ^2 x}=10 \\\\ &\Rightarrow \frac{9}{9^{\tan ^2 x}}+9^{\tan ^2 x}=10 \end{aligned}$ Let $9^{\tan ^2 x}=t$ $\frac{9}{t}+t=10 \Rightarrow t^2-10 t+9=0 \Rightarrow t=9 \text { or } 1$ If $t=9,9^{\tan ^2 x}=9 \Rightarrow x= \pm \frac{\pi}{4}$ If $t=1,9^{\tan ^2 x}=1 \Rightarrow x=0$ Also, $\begin{aligned} \beta & =\sum_{x \in S} \tan ^2\left(\frac{x}{3}\right) \\\\ & =\tan ^2\left(\frac{0}{3}\right)+\tan ^2\left(\frac{\pi}{12}\right)+\tan ^2\left(-\frac{\pi}{12}\right) \\\\ & =0+2 \tan ^2 \frac{\pi}{12}=2[2-\sqrt{3}]^2 \\\\ & =2[4+3-4 \sqrt{3}]=14-8 \sqrt{3} \end{aligned}$ $\text { So, now } \frac{1}{6}(\beta-14)^2=\frac{1}{6}[14-8 \sqrt{3}-14]^2=\frac{1}{6} \times 64 \times 3=32$