The maximum value of $$f(x) = \begin{vmatrix} \sin^2 x & 1 + \cos^2 x & \cos 2x \\ 1 + \sin^2 x & \cos^2 x & \cos 2x \\ \sin^2 x & \cos^2 x & \sin 2x \end{vmatrix}$$, $$x \in R$$ is:

Let $$s = \sin x$$ and $$c = \cos x$$.
Then $$\sin^2 x = s^2,\; \cos^2 x = c^2,\; \cos 2x = c^2 - s^2,\; \sin 2x = 2sc$$.

The determinant becomes

$$f(x)=\begin{vmatrix}s^2 & 1+c^2 & c^2-s^2 \\ 1+s^2 & c^2 & c^2-s^2 \\ s^2 & c^2 & 2sc\end{vmatrix}$$.

Apply the row operations $$R_2 \rightarrow R_2 - R_1$$ and $$R_3 \rightarrow R_3 - R_1$$:

$$f(x)=\begin{vmatrix}s^2 & 1+c^2 & c^2-s^2 \\ 1 & -1 & 0 \\ 0 & -1 & 2sc-c^2+s^2\end{vmatrix}$$.

Expand using the first row:
First form the cross product of the second and third rows: $$[1,-1,0]\times[0,-1,2sc-c^2+s^2]=(-t,\,-t,\,-1)$$ where $$t=2sc-c^2+s^2$$.

Dotting this with the first row gives

$$f(x)=s^2(-t)+(1+c^2)(-t)+(c^2-s^2)(-1)$$.

Since $$s^2+c^2=1$$, we get $$s^2+1+c^2=2$$, hence

$$f(x)=-2t-(c^2-s^2)$$.

Substituting $$t=2sc-(c^2-s^2)$$:

$$f(x)=-2\left(2sc-(c^2-s^2)\right)-(c^2-s^2)=-4sc+(c^2-s^2)$$.

Using $$c^2-s^2=\cos 2x$$ and $$2sc=\sin 2x$$, we obtain

$$f(x)=\cos 2x-2\sin 2x$$.

Put $$y=2x$$ (so $$y\in\mathbb{R}$$). Then

$$f(x)=g(y)=\cos y-2\sin y$$.

Write this as a single sine function.
The general form $$a\cos y+b\sin y$$ has amplitude $$\sqrt{a^2+b^2}$$.
Here $$a=1,\;b=-2\Rightarrow \sqrt{a^2+b^2}=\sqrt{1+4}=\sqrt{5}$$.

Therefore $$-\sqrt{5}\le g(y)\le\sqrt{5}$$, so the maximum value is $$\sqrt{5}$$.

Hence the maximum value of $$f(x)$$ is $$\sqrt{5}$$, which corresponds to Option C.