Let $$S$$ be the set of all values of $$\theta \in [-\pi, \pi]$$ for which the system of linear equations
$$x + y + \sqrt{3}z = 0$$
$$-x + (\tan\theta)y + \sqrt{7}z = 0$$
$$x + y + (\tan\theta)z = 0$$
has non-trivial solution. Then $$\frac{120}{\pi}\sum_{\theta \in S} \theta$$ is equal to

For a non-trivial solution, the determinant of the coefficient matrix must be zero:

$$\begin{vmatrix} 1 & 1 & \sqrt{3} \\ -1 & \tan\theta & \sqrt{7} \\ 1 & 1 & \tan\theta \end{vmatrix} = 0$$

Expanding along the first row:

$$1(\tan^2\theta - \sqrt{7}) - 1(-\tan\theta - \sqrt{7}) + \sqrt{3}(-1 - \tan\theta) = 0$$

$$\tan^2\theta - \sqrt{7} + \tan\theta + \sqrt{7} - \sqrt{3} - \sqrt{3}\tan\theta = 0$$

$$\tan^2\theta + \tan\theta - \sqrt{3}\tan\theta - \sqrt{3} = 0$$

$$\tan^2\theta + (1 - \sqrt{3})\tan\theta - \sqrt{3} = 0$$

This factors as:

$$(\tan\theta + 1)(\tan\theta - \sqrt{3}) = 0$$

Check: $$\tan\theta \cdot (-\sqrt{3}) + 1 \cdot \tan\theta = \tan\theta(1 - \sqrt{3})$$ ✓ and $$1 \cdot (-\sqrt{3}) = -\sqrt{3}$$ ✓

So $$\tan\theta = -1$$ or $$\tan\theta = \sqrt{3}$$.

In $$[-\pi, \pi]$$:

$$\tan\theta = -1$$: $$\theta = -\frac{\pi}{4}, \frac{3\pi}{4}$$

$$\tan\theta = \sqrt{3}$$: $$\theta = \frac{\pi}{3}, -\frac{2\pi}{3}$$

Sum = $$-\frac{\pi}{4} + \frac{3\pi}{4} + \frac{\pi}{3} - \frac{2\pi}{3} = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$$

$$\frac{120}{\pi} \times \frac{\pi}{6} = 20$$

The correct answer is Option 1: 20.