Since  $\mathrm{A}.\mathrm{M}\mathrm{ }\ge \mathrm{G}.\mathrm{M}$

$\frac{x+y+z}{3} \ge {\left(xyz\right)}^{\frac{1}{3}}$

$x+y+z \ge 3{\left(xyz\right)}^{\frac{1}{3}}$

Now, $\left|\begin{array}{c}x\\ 1\\ 1\end{array} \begin{array}{c}1\\ y\\ 1\end{array} \begin{array}{c}1\\ 1\\ z\end{array}\right|=xyz-\left(x+y+z\right)+2$

Now determinant is non-negative

$\Rightarrow xyz-\left(x+y+z\right)+2\ge 0$

$\Rightarrow xyz-\left(3\right){\left(xyz\right)}^{\frac{1}{3}}+2 \ge 0$

Let $xyz=t^3$

So, $t^3-3t+2 \ge 0$

$$\begin{gathered} \Rightarrow \left(t+2\right)(t^2-2t+1)\ge 0 \\ \Rightarrow \left(t+2\right){(t-1)}^2\ge 0 \end{gathered}$$

$\Rightarrow t\ge -2$

$\Rightarrow t^3\ge -8$

$\Rightarrow$ Least value of $xyz= -8$