The number of $\theta \in (0,4\pi )$ for which the system of linear equations $3(\sin 3\theta )x-y+z=2, 3(\cos 2\theta )x+4y+3z=3$, $6x+7y+7z=9$ has no solution is :

The system of equation has no solution.

$\begin{array}{l}\mathrm{D}=\left|\begin{array}{ccc}3\sin 3\theta & -1& 1\\ 3\cos 2\theta & 4& 3\\ 6& 7& 7\end{array}\right|=0\\ 21\sin 3\theta +42\cos 2\theta -42=0\\ \sin 3\theta +2\cos 2\theta -2=0\\ 3\sin \theta -4{\sin }^3\theta -2\left(2{\sin }^2\theta \right)=0\\ \sin \theta (4{\sin }^2\theta +4\sin \theta -3)=0\\ \sin \theta =0,4{\sin }^2\theta +6\sin \theta -2\sin \theta -3=0\\ \sin \theta =0,2\sin \theta (2\sin \theta +3)-(2\sin \theta +3)=0\\ \sin \theta =0,\sin \theta =\frac{1}{2},\\ \theta =\left\{\frac{\mathrm{\pi }}{6},\frac{5\mathrm{\pi }}{6},\frac{13\mathrm{\pi }}{6},\frac{17\mathrm{\pi }}{6},\mathrm{\pi },2\mathrm{\pi },3\mathrm{\pi }\right\}\end{array}$

Number of solution is 7 in $(0,4\pi )$