Let $D= \begin{vmatrix}1&sin \theta & 1\\-sin \theta & 1& sin \theta \\-1& -sin \theta & 1\end {vmatrix}; 0≤\theta ≤ 2\pi , $ then

$D \in [2, 4]$

The correct answer is option (3) : $D \in [2, 4]$

We have,

$D=\begin{vmatrix}1&sin \theta & 1\\-sin \theta & 1& sin \theta \\-1& -sin \theta & 1\end {vmatrix}$

$⇒D=\begin{vmatrix}2 &sin \theta & 1\\0 & 1 & sin \theta \\0 & -sin \theta & 1\end {vmatrix}$    [Applying $C_1→C_1+C_3$]

$⇒D= 2(1+ sin^2 \theta )$

Now, $0≤sin^2\theta ≤ 1$

$⇒2≤2(1+sin^2 \theta ) ≤ 4 $

$⇒ D \in [2, 4]$