The value of the determinant $$ \left|\begin{array}{ccc} \sin \theta & \cos \theta & \sin 2 \theta \\ \sin [\theta+(2 \pi / 3)] & \cos [\theta+(2 \pi / 3)] & \sin [2 \theta+(4 \pi / 3)] \\ \sin [\theta-(2 \pi / 3)] & \cos [\theta-(2 \pi / 3)] & \sin [2 \theta-(4 \pi / 3)] \end{array}\right| $$ is (a) 0 (b) \(2 \sin \theta\) \(1-\sin 2 \theta \quad\) (d) \(-2 \cos \theta\)

We can rewrite the trigonometric functions in the second and third row using the identities \(\sin(a + b)\) and \(\cos(a + b)\). The given determinant is: $$ \left|\begin{array}{ccc} \sin \theta & \cos \theta & \sin 2 \theta \\\ \sin [\theta+(2 \pi / 3)] & \cos [\theta+(2 \pi / 3)] & \sin [2 \theta+(4 \pi / 3)] \\\ \sin [\theta-(2 \pi / 3)] & \cos [\theta-(2 \pi / 3)] & \sin [2 \theta-(4 \pi / 3)] \end{array}\right| $$ After applying the identities, we get: $$ \left|\begin{array}{ccc} \sin \theta & \cos \theta & \sin 2 \theta \\\ \sin \theta \cos(2\pi/3) + \cos \theta \sin(2\pi/3) & \cos \theta \cos(2\pi/3) - \sin \theta \sin(2\pi/3) & \sin 2 \theta \cos(4\pi/3) + \cos 2 \theta \sin(4\pi/3) \\ \sin \theta \cos(-2\pi/3) + \cos \theta \sin(-2\pi/3) & \cos \theta \cos(-2\pi/3) - \sin \theta \sin(-2\pi/3) & \sin 2 \theta \cos(-4\pi/3) + \cos 2 \theta \sin(-4\pi/3) \end{array}\right| $$

Now, we simplify the determinant by evaluating the trigonometric functions at the given angles: $$ \left|\begin{array}{ccc} \sin \theta & \cos \theta & \sin 2 \theta \\\ -\frac{1}{2}\sin \theta + \frac{\sqrt{3}}{2}\cos \theta & -\frac{\sqrt{3}}{2}\sin \theta -\frac{1}{2}\cos\theta & -\sin 2 \theta \\ \frac{1}{2}\sin \theta - \frac{\sqrt{3}}{2}\cos \theta & \frac{\sqrt{3}}{2}\sin \theta - \frac{1}{2}\cos\theta & \sin 2 \theta \end{array}\right| $$

We will expand the determinant along the first row: $$ \begin{aligned} &=\sin \theta \cdot \left|\begin{array}{cc} -\frac{\sqrt{3}}{2}\sin \theta -\frac{1}{2}\cos\theta & -\sin 2 \theta \\ \frac{\sqrt{3}}{2}\sin \theta - \frac{1}{2}\cos\theta & \sin 2 \theta \end{array}\right| - \cos \theta \cdot \left|\begin{array}{cc} -\frac{1}{2}\sin \theta + \frac{\sqrt{3}}{2}\cos \theta & -\sin 2 \theta \\ \frac{1}{2}\sin \theta - \frac{\sqrt{3}}{2}\cos \theta & \sin 2 \theta \end{array}\right| + \sin 2 \theta \cdot \left|\begin{array}{cc} -\frac{1}{2}\sin \theta + \frac{\sqrt{3}}{2}\cos \theta & -\frac{\sqrt{3}}{2}\sin \theta -\frac{1}{2}\cos\theta \\ \frac{1}{2}\sin \theta - \frac{\sqrt{3}}{2}\cos \theta & \frac{\sqrt{3}}{2}\sin \theta - \frac{1}{2}\cos\theta \end{array}\right| \end{aligned} $$

Compute the values of the smaller determinants: $$ \begin{aligned} &=\sin \theta \cdot [-\frac{3}{4}\sin^2 \theta - \frac{1}{4}\cos^2\theta -\frac{\sqrt{3}}{2}\sin^2 2 \theta] - \cos \theta \cdot [-\frac{3}{4}\cos^2 \theta - \frac{1}{4}\sin^2\theta + \frac{\sqrt{3}}{2}\sin 2 \theta \cos 2 \theta] \\ + & \quad \sin 2 \theta \cdot[-\frac{3}{4}\sin \theta \cos \theta + \frac{1}{4}\sin \theta \cos\theta +\frac{\sqrt{3}}{2}\sin^2\theta - \frac{\sqrt{3}}{2}\cos^2\theta] \end{aligned} $$

Simplify the expression and combine like terms: $$ \begin{aligned} &=-\frac{1}{2}\sin \theta - \frac{\sqrt{3}}{2}\sin \theta \cos \theta - \frac{\sqrt{3}}{2}\sin 2 \theta \cos 2 \theta + \frac{1}{2} \sin 2 \theta \cos \theta + \frac{\sqrt{3}}{2} \sin^2 \theta \cos \theta \\ \end{aligned} $$ Looking at the given options, none exactly matches the obtained expression. However, we can notice that our simplified expression equals 0 when \(\sin \theta = 0\) or when \(\cos \theta = 0\). This might suggest some errors in the calculation or that the correct choice is (a) 0. However, without more context, we cannot be certain.