Animated Solution for Mathematics - Matrices and Determinants: Let $S$ be the set of all values of $\theta \in [-\pi ,\pi ]$ for which the system of linear equations $x+y+\sqrt{3}z=0$, $-x+\left(\tan \theta \right)y+\sqrt{7}z=0$, $x+y+\left(\tan \theta \right)z=0$ has non-trivial solution. Then $120{\sum }_{\theta \in S}\theta$ is equal to

Imagine you are standing before a system of three homogeneous linear equations. In the world of JEE Advanced, these systems are geometric entities where a non-trivial solution implies that the three planes intersect in a line rather than at a single point.

The key to unlocking this is the determinant of the coefficient matrix. By setting the determinant $|A|=0$, we force the system to be dependent, which is the necessary condition for non-trivial solutions to exist.

Let us construct our determinant from the coefficients:

Expanding this along the first row, we obtain:

As we simplify, the $\sqrt{7}$ terms cancel out, leaving us with a quadratic equation in $\tan \theta$:

Factoring this expression, we find:

Now, we solve for $\theta$ in the interval $[-\pi ,\pi ]$. For $\tan \theta =\sqrt{3}$, we look at the 1st and 3rd quadrants, yielding $\theta =\frac{\pi }{3}$ and $\theta =-\frac{2\pi }{3}$.

For $\tan \theta =-1$, we look at the 2nd and 4th quadrants, yielding $\theta =\frac{3\pi }{4}$ and $\theta =-\frac{\pi }{4}$.

Summing these values, we get:

Finally, calculating the requested value:

The beauty of this problem lies in how the complex-looking coefficients simplify into a clean, solvable quadratic, rewarding your patience and precision. The final answer is 20.