Let d ∈ R, and A = [ cases - 2 & 4 + d & (sinθ) - 2 1 & (sinθ) + 2...

\left|A\right| = \left|$\begin{cases}- 2 & 4 + d & \left(sin\theta\right) - 2 \\ 1 & \left(sin\theta\right) + 2 & d \\ 5 & \left(2sin\theta\right) - d & \left(- sin\theta\right) + 2 + 2d\end{cases}$\right|$</p><p>${R}_{3} \to {R}_{3} - 2{R}_{2} + {R}_{1}$</p><p>$= \left|$\begin{cases}- 2 & 4 + d & \left(sin\theta\right) - 2 \\ 1 & \left(sin\theta\right) + 2 & d \\ 1 & 0 & 0\end{cases}$\right|$</p><p>$= 1\left\{\left(4 + d\right)d - \left(sin\theta + 2\right)\left(sin\theta - 2\right)\right\}$</p><p>$= 4d + {d}^{2} - \sin^{2}\theta + 4 = {\left(d + 2\right)}^{2} - \sin^{2}\theta$</p><p>We know that$0 \le \sin^{2}\theta \le 1$</p><p>Given, the minimum value of \left|A\right| = 8,$ which is possible, when $\sin^{2}\theta = 1$

$⇒ {\left(d + 2\right)}^{2} = 9$

$⇒ d + 2 = \pm 3$

$⇒ d = 1$ or $d = - 5 .$