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(B) For the system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.$\Delta = \begin{vmatrix}

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$4$

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(B) For the system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.$\Delta = \begin{vmatrix} \sin 3 	heta & -1 & 1 \ \cos 2 	heta & 4 & 3 \ 2 & 7 & 7 \end{vmatrix} = 0$Expanding along the first row:$\sin 3 	heta (4 	imes 7 - 3 	imes 7) - (-1) (\cos 2 	heta 	imes 7 - 3 	imes 2) + 1 (\cos 2 	heta 	imes 7 - 4 	imes 2) = 0$$\sin 3 	heta (28 - 21) + (7 \cos 2 	heta - 6) + (7 \cos 2 	heta - 8) = 0$$7 \sin 3 	heta + 14 \cos 2 	heta - 14 = 0$$\sin 3 	heta + 2 \cos 2 	heta - 2 = 0$Using $\cos 2 	heta = 1 - 2 \sin^2 	heta$ and $\sin 3 	heta = 3 \sin 	heta - 4 \sin^3 	heta$:$3 \sin 	heta - 4 \sin^3 	heta + 2(1 - 2 \sin^2 	heta) - 2 = 0$$3 \sin 	heta - 4 \sin^3 	heta + 2 - 4 \sin^2 	heta - 2 = 0$$-4 \sin^3 	heta - 4 \sin^2 	heta + 3 \sin 	heta = 0$$-\sin 	heta (4 \sin^2 	heta + 4 \sin 	heta - 3) = 0$$-\sin 	heta (2 \sin 	heta - 1)(2 \sin 	heta + 3) = 0$This gives $\sin 	heta = 0$,$\sin 	heta = 1/2$,or $\sin 	heta = -3/2$ (impossible).For $\sin 	heta = 0$,$	heta = 0, \pi$ (in $[0, 2\pi)$).For $\sin 	heta = 1/2$,$	heta = \pi/6, 5\pi/6$ (in $[0, 2\pi)$).Thus,there are $4$ principal values of $	heta$.

If $x, y, z$ are not all simultaneously equal to zero,satisfying the system of equations
$(\sin 3 \theta) x - y + z = 0$
$(\cos 2 \theta) x + 4 y + 3 z = 0$
$2 x + 7 y + 7 z = 0$
then the number of principal values of $\theta$ is

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If $x, y, z$ are not all simultaneously equal to zero,satisfying the system of equations$(\sin 3 \theta) x - y + z = 0$$(\cos 2 \theta) x + 4 y + 3 z = 0$$2 x + 7 y + 7 z = 0$then the number of principal values of $\theta$ is