Let $$f(x) = \begin{vmatrix} 1+\sin^2 x & \cos^2 x & \sin 2x \\ \sin^2 x & 1+\cos^2 x & \sin 2x \\ \sin^2 x & \cos^2 x & 1+\sin 2x \end{vmatrix}$$, $$x \in \left[\frac{\pi}{6}, \frac{\pi}{3}\right]$$. If $$\alpha$$ and $$\beta$$ respectively are the maximum and the minimum values of $$f$$, then

Given: $$f(x) = \begin{vmatrix} 1+\sin^2 x & \cos^2 x & \sin 2x \\ \sin^2 x & 1+\cos^2 x & \sin 2x \\ \sin^2 x & \cos^2 x & 1+\sin 2x \end{vmatrix}$$ for $$x \in [\pi/6, \pi/3]$$.

Apply row operations $$R_1 \to R_1 - R_3$$ and $$R_2 \to R_2 - R_3$$. These operations do not change the value of the determinant.

$$R_1 \to (1+\sin^2 x - \sin^2 x, \; \cos^2 x - \cos^2 x, \; \sin 2x - 1 - \sin 2x) = (1, 0, -1)$$

$$R_2 \to (\sin^2 x - \sin^2 x, \; 1+\cos^2 x - \cos^2 x, \; \sin 2x - 1 - \sin 2x) = (0, 1, -1)$$

$$R_3$$ remains $$(\sin^2 x, \cos^2 x, 1+\sin 2x)$$.

Expand the determinant along $$R_1$$:

$$f(x) = 1 \cdot \begin{vmatrix} 1 & -1 \\ \cos^2 x & 1+\sin 2x \end{vmatrix} - 0 + (-1) \cdot \begin{vmatrix} 0 & 1 \\ \sin^2 x & \cos^2 x \end{vmatrix}$$

First minor: $$1(1+\sin 2x) - (-1)(\cos^2 x) = 1 + \sin 2x + \cos^2 x$$.

Second minor: $$0(\cos^2 x) - 1(\sin^2 x) = -\sin^2 x$$.

$$f(x) = 1 + \sin 2x + \cos^2 x + \sin^2 x = 1 + \sin 2x + 1 = 2 + \sin 2x$$

Find $$\alpha$$ (maximum) and $$\beta$$ (minimum) of $$f(x) = 2 + \sin 2x$$ on $$[\pi/6, \pi/3]$$.

For $$x \in [\pi/6, \pi/3]$$, we have $$2x \in [\pi/3, 2\pi/3]$$.

On $$[\pi/3, 2\pi/3]$$: $$\sin$$ achieves maximum at $$2x = \pi/2$$, i.e., $$x = \pi/4$$, giving $$\sin(\pi/2) = 1$$.

At endpoints: $$\sin(\pi/3) = \sin(2\pi/3) = \frac{\sqrt{3}}{2}$$.

Maximum: $$\alpha = 2 + 1 = 3$$.

Minimum: $$\beta = 2 + \frac{\sqrt{3}}{2}$$.

Check Option A: $$\beta^2 - 2\sqrt{\alpha}$$.

$$\beta^2 = \left(2 + \frac{\sqrt{3}}{2}\right)^2 = 4 + 2\sqrt{3} + \frac{3}{4} = \frac{19}{4} + 2\sqrt{3}$$

$$2\sqrt{\alpha} = 2\sqrt{3}$$

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

The correct answer is Option A: $$\beta^2 - 2\sqrt{\alpha} = \frac{19}{4}$$.