Given:

$f\left(x\right)=\left|\begin{array}{ccc}1+{\sin }^{2}x& {\cos }^{2}x& \sin 2x\\ {\sin }^{2}x& 1+{\cos }^{2}x& \sin 2x\\ {\sin }^{2}x& {\cos }^{2}x& 1+\sin 2x\end{array}\right|$

Applying $C_1\to C_1+C_2+C_3$

$f\left(x\right)=\left|\begin{array}{ccc}2+\sin 2x& {\cos }^{2}x& \sin 2x\\ 2+\sin 2x& 1+{\cos }^{2}x& \sin 2x\\ 2+\sin 2x& {\cos }^{2}x& 1+\sin 2x\end{array}\right|$

$\Rightarrow f\left(x\right)=\left(2+\sin 2x\right)\left|\begin{array}{ccc}1& {\cos }^{2}x& \sin 2x\\ 1& 1+{\cos }^{2}x& \sin 2x\\ 1& {\cos }^{2}x& 1+\sin 2x\end{array}\right|$

Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$

$f\left(x\right)=\left(2+\sin 2x\right)\left|\begin{array}{ccc}1& {\cos }^{2}x& \sin 2x\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|$

$\Rightarrow f\left(x\right)=\left(2+\sin 2x\right)\left(1\right)=2+\sin 2x$

Now, for 

$x\in \left[\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{3}\right]$

$\Rightarrow 2x\in \left[\frac{\mathrm{\pi }}{3},\frac{2\mathrm{\pi }}{3}\right]$

$\Rightarrow \sin 2x\in \left[\frac{\sqrt{3}}{2},1\right]$

$\Rightarrow 2+\sin 2x\in \left[2+\frac{\sqrt{3}}{2},3\right]$

Hence,

$\beta =2+\frac{\sqrt{3}}{2}$

$\alpha =3$

So,

${\beta }^2-2\alpha =4+\frac{3}{4}+2\sqrt{3}-2\sqrt{3}=\frac{19}{4}$