Let d ∈ R, and A= [-2, 4+d, ( sin θ)-2; 1, ( sin θ)+2, d; 5, (2 sin θ)-d, (- sin θ)+2+2 d], θ ∈[0,2 π]. If the minimum value of det ( A ) is 8, then…

Let $\displaystyle d \in \mathbf{R}$, and $$A=\left[\begin{array}{ccc} -2 & 4+d & (\sin \theta)-2 \\ 1 & (\sin \theta)+2 & d \\ 5 & (2 \sin \theta)-d & (-\sin \theta)+2+2 d \end{array}\right], $$ $\displaystyle \theta \in[0,2 \pi]$. If the minimum value of $\displaystyle \operatorname{det}(\mathrm{A})$ is $\displaystyle 8$, then a value of d is :