Choose the correct answer.

Answer: D

$A=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]$

$\therefore|A|=1\left(1+\sin ^{2} \theta\right)-\sin \theta(-\sin \theta+\sin \theta)+1\left(\sin ^{2} \theta+1\right)$

$=1+\sin ^{2} \theta+\sin ^{2} \theta+1$

$=2+2 \sin ^{2} \theta$

$=2\left(1+\sin ^{2} \theta\right)$

Now, $0 \leq \theta \leq 2 \pi$

$\Rightarrow-1 \leq \sin \theta \leq 1$

$\Rightarrow 0 \leq \sin ^{2} \theta \leq 1$

$\Rightarrow 1 \leq 1+\sin ^{2} \theta \leq 2$

$\Rightarrow 2 \leq 2\left(1+\sin ^{2} \theta\right) \leq 4$

$\therefore \operatorname{Det}(A) \in[2,4]$

The correct answer is D.