The determinant is given by:
\[ D = \left| \begin{matrix} 1 & 1 & \sqrt{3} \\ -1 & \tan \theta & \sqrt{7} \\ 1 & 1 & \tan \theta \end{matrix} \right| = 0 \]
Simplifying the determinant expression: \[ \tan^2 \theta - (\sqrt{3} - 1) - \sqrt{3} = 0 \]
Solving for \( \tan \theta \): \[ \tan \theta = \sqrt{3}, -1 \]
Therefore, the possible values for \( \theta \) are: \[ \theta = \left\{ \frac{\pi}{3}, -\frac{2\pi}{3}, -\frac{\pi}{4}, \frac{3\pi}{4} \right\} \]
Now, calculating the sum of angles: \[ \frac{120}{\pi} \left( \sum \theta \right) = \frac{120}{\pi} \times \frac{\pi}{6} = 20 \quad (\text{Option 2}) \]