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

  • A
    $32$
  • B
    $8$
  • C
    $64$
  • D
    $16$

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