$\left|A\right|=\left|\begin{array}{ccc}-2& 4+d& \left(\mathrm{s}\mathrm{i}\mathrm{n}\theta \right)-2\\ 1& \left(\sin \theta \right)+2& d\\ 5& \left(2\sin \theta \right)-d& \left(-\sin \theta \right)+2+2d\end{array}\right|$

$R_3\to R_3-2R_2+R_1$

$=\left|\begin{array}{ccc}-2& 4+d& \left(\mathrm{s}\mathrm{i}\mathrm{n}\theta \right)-2\\ 1& \left(\sin \theta \right)+2& d\\ 1& 0& 0\end{array}\right|$

$=1\left\{\left(4+d\right)d-\left(\sin \theta +2\right)\left(\sin \theta -2\right)\right\}$

$=4d+d^2-{\sin }^2\theta +4={\left(d+2\right)}^2-{\sin }^2\theta$

We know that $0\le {\sin }^2\theta \le 1$

Given, the minimum value of $\left|A\right|=8,$ which is possible, when ${\sin }^2\theta =1$

$\Rightarrow {\left(d+2\right)}^2=9$

$\Rightarrow d+2=\pm 3$

$\Rightarrow d=1$ or $d=-5.$