If $$f(x) = \begin{vmatrix} 2\cos^4 x & 2\sin^4 x & 3 + \sin^2 2x \\ 3 + 2\cos^4 x & 2\sin^4 x & \sin^2 2x \\ 2\cos^4 x & 3 + 2\sin^4 x & \sin^2 2x \end{vmatrix}$$ then $$\frac{1}{5}f'(0)$$ is equal to ________.

Given the determinant function $$f(x) = \begin{vmatrix} 2\cos^4 x & 2\sin^4 x & 3 + \sin^2 2x \\ 3 + 2\cos^4 x & 2\sin^4 x & \sin^2 2x \\ 2\cos^4 x & 3 + 2\sin^4 x & \sin^2 2x \end{vmatrix}$$, find $$\frac{1}{5}f'(0)$$.

Let $$a = 2\cos^4 x$$, $$b = 2\sin^4 x$$, $$c = \sin^2 2x$$.

Note that $$a + b + c = 2\cos^4 x + 2\sin^4 x + \sin^2 2x = 2(\cos^4 x + \sin^4 x) + 4\sin^2 x\cos^2 x$$

$$= 2(\cos^2 x + \sin^2 x)^2 - 4\sin^2 x\cos^2 x + 4\sin^2 x\cos^2 x = 2$$.

So $$a + b + c = 2$$ and the third column of row 1 is $$3 + c = 3 + c$$, while the first column entries are $$a, 3+a, a$$.

Apply $$R_2 \to R_2 - R_1$$ and $$R_3 \to R_3 - R_1$$:

$$\begin{vmatrix} a & b & 3+c \\ 3 & 0 & -3 \\ 0 & 3 & -3 \end{vmatrix}$$

Expanding: $$a(0-(-9)) - b(-9-0) + (3+c)(9-0)$$

$$= 9a + 9b + 9(3+c) = 9(a + b + 3 + c) = 9(2 + 3) = 45$$.

So $$f(x) = 45$$ for all $$x$$ (constant function).

Since $$f(x) = 45$$ is constant, $$f'(x) = 0$$ for all $$x$$.

$$\frac{1}{5}f'(0) = \frac{0}{5} = 0$$.

The correct answer is Option A: 0.