Let $Δ=\begin{vmatrix} 1  & sin \theta & 1\\-sin\theta & 1 & sin \theta \\-1 & -sin \theta & 1 \end{vmatrix}$ then Δ lies in the interval :

[2, 4]

The correct answer is Option (2) → [2, 4]

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

$=1(1+\sin θ)+\sin θ(-\sin θ+\sin θ)+1(1+\sin^2 θ)$

$=2+2\sin^2 θ$

as $0≤\sin^2 θ≤1⇒1≤1+\sin^2 θ≤2$

so $2≤2(1+\sin^2 θ)≤4$

so $Δ∈[2, 4]$