Find the value of $\theta$ satisfying $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1&{\sin 3\theta }\\{ - 4}&3&{\cos 2\theta }\\7&{ - 7}&{ - 2}\end{array}} \right] = 0$

We have, $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1&{\sin 3\theta }\\{ - 4}&3&{\cos 2\theta }\\7&{ - 7}&{ - 2}\end{array}} \right| = 0$

$\Rightarrow$ $\left| {\begin{array}{cccccccccccccccccccc}0&1&{\sin 3\theta }\\{ - 7}&3&{\cos 2\theta }\\{14}&{ - 7}&{ - 2}\end{array}} \right| = 0$

$\Rightarrow$ $7\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&1&{\sin 3\theta }\\{ - 1}&3&{\cos 2\theta }\\2&{ - 7}&{ - 2}\end{array}} \right| = 0$

[taking 7 common from ${C_1}$]

$\Rightarrow$ $7[0 - 1(2 - 2\cos 2\theta ) + \sin 3\theta (7 - 6)] = 0$ [expanding along ${R_1}$]

$\Rightarrow$ $7[ - 2(1 - \cos 2\theta ) + \sin 3\theta ] = 0$

$\Rightarrow$ $- 14 + 14\cos 2\theta + 7\sin 3\theta = 0$

$\Rightarrow$ $14\cos 2\theta + 7\sin 3\theta = 14$

$\Rightarrow$ $14\left( {1 - 2{{\sin }^2}\theta } \right) + 7\left( {3\sin \theta - 4{{\sin }^3}\theta } \right) = 14$

$\Rightarrow$ $- 28{\sin ^2}\theta + 14 + 21\sin \theta - 28{\sin ^3}\theta = 14$

$\Rightarrow$ $- 28{\sin ^2}\theta - 28{\sin ^3}\theta + 21\sin \theta = 0$

$\Rightarrow$ $28{\sin ^3}\theta + 28{\sin ^2}\theta - 21\sin \theta = 0$

$\Rightarrow$ $4{\sin ^3}\theta + 4{\sin ^2}\theta - 3\sin \theta = 0$

$\Rightarrow$ $\sin \theta \left( {4{{\sin }^2}\theta + 4\sin \theta - 3} \right) = 0$

$\Rightarrow$ Either $\sin \theta = 0$,

$\Rightarrow$ $\theta = n\pi$ or $4{\sin ^2}\theta + 4\sin \theta - 3 = 0$

$\sin \theta = \frac{{ - 4 \pm \sqrt {16 + 48} }}{8} = \frac{{ - 4 \pm \sqrt {64} }}{8}$

$= \frac{{ - 4 \pm 8}}{8} = \frac{4}{8},\frac{{ - 12}}{8}$

$\sin \theta = \frac{1}{2},\frac{{ - 3}}{2}$

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

then
$\theta = n\pi + {( - 1)^n}\frac{\pi }{6}$
Hence, $\sin \theta = \frac{{ - 3}}{2}$

[not possible because $- 1 \le \sin \theta \le 1$]