JEE Main 2023 — Determinants Question with Solution

Given,

$S$ 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,

So, by condition of non-trivial solution we get,

$\left|\begin{array}{ccc}1& 1& \sqrt{3}\\ -1& \tan \theta & \sqrt{7}\\ 1& 1& \tan \theta \end{array}\right|=0$

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

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

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

$\Rightarrow \tan \theta =\sqrt{3},-1$

$\Rightarrow \theta =\frac{\pi }{3},\frac{-2\pi }{3},\frac{-\pi }{4},\frac{3\pi }{4}$

Hence,

$\frac{120}{\pi }\sum \theta =120\left(\frac{1}{3}-\frac{2}{3}-\frac{1}{4}+\frac{3}{4}\right)=20$