The number of values of $\theta \in (0, \pi ) $ for which the system of linear equations

$x+3y + 7 z=0 $

$-x+ 4y + 7z= 0 $

$(sin 3\theta  ) x+ (cos 2 \theta ) y + 2z= 0 $

has a non-trivial solution, is

two

The correct answer is option (4) : two

The given homogeneous system of equations will have non-trivial solutions, if

$\begin{vmatrix}1 & 3 & 7\\-1 & 4 & 7\\sin 3\theta  & cos 2\theta  & 2\end{vmatrix}=0$

$⇒\begin{vmatrix}1 & 0 & 0\\-1 & 7 & 14\\sin 3\theta  & cos 2\theta -3sin 3\theta  & 2-7 sin 3\theta \end{vmatrix}=0$

Applying $C_2→C_2-3C_1,C_3 → C_3 -7C_1$

$⇒14-49sin 3\theta - 14 cos 2 \theta + 42 sin 3 \theta = 0 $

$⇒28sin^2 \theta - 7 (3sin \theta - 4 sin^3 \theta ) = 0 $

$⇒28sin^3\theta + 28 sin^2 \theta - 21 sin \theta = 0 $

$⇒ 7 sin \theta (4 sin^2 \theta + 4 sin \theta - 3) = 0 $

$⇒ 7 sin \theta ( 2 sin \theta + 3) (2 sin \theta - 1) = 0 $

$⇒2 sin \theta - 1= 0 $         $[∵ \theta \in (0, \pi ) ∴sin \theta ≠0\, and\, \, sin \theta ≠-\frac{3}{2}]$

$⇒sin \theta = \frac{1}{2}⇒\theta =\frac{\pi }{6}, \frac{5\pi }{6}$

Hence, there are two values of $\theta $ in $(0, \pi ).$