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(C) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$D = 0$.$\begin{vmatri

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$20$

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(C) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$D = 0$.$\begin{vmatrix} 1 & 1 & \sqrt{3} \ -1 & 	an 	heta & \sqrt{7} \ 1 & 1 & 	an 	heta \end{vmatrix} = 0$Expanding along the first row:$1(	an^2 	heta - \sqrt{7}) - 1(-	an 	heta - \sqrt{7}) + \sqrt{3}(-1 - 	an 	heta) = 0$$	an^2 	heta - \sqrt{7} + 	an 	heta + \sqrt{7} - \sqrt{3} - \sqrt{3} 	an 	heta = 0$$	an^2 	heta + (1 - \sqrt{3}) 	an 	heta - \sqrt{3} = 0$$(	an 	heta - \sqrt{3})(	an 	heta + 1) = 0$Thus,$	an 	heta = \sqrt{3}$ or $	an 	heta = -1$.For $	an 	heta = \sqrt{3}$ and $	heta \in [-\pi, \pi]$,$	heta = \frac{\pi}{3}, -\frac{2\pi}{3}$.For $	an 	heta = -1$ and $	heta \in [-\pi, \pi]$,$	heta = \frac{3\pi}{4}, -\frac{\pi}{4}$.The sum of all values in $S$ is $\sum_{	heta \in S} 	heta = \frac{\pi}{3} - \frac{2\pi}{3} + \frac{3\pi}{4} - \frac{\pi}{4} = -\frac{\pi}{3} + \frac{2\pi}{4} = -\frac{\pi}{3} + \frac{\pi}{2} = \frac{\pi}{6}$.Therefore,$\frac{120}{\pi} \sum_{	heta \in S} 	heta = \frac{120}{\pi} 	imes \frac{\pi}{6} = 20$.

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 a non-trivial solution. Then $\frac{120}{\pi} \sum_{\theta \in S} \theta$ is equal to

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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 a non-trivial solution. Then $\frac{120}{\pi} \sum_{\theta \in S} \theta$ is equal to