If $$f(x) = \begin{vmatrix} x^3 & 2x^2+1 & 1+3x \\ 3x^2+2 & 2x & x^3+6 \\ x^3-x & 4 & x^2-2 \end{vmatrix}$$ for all $$x \in \mathbb{R}$$, then $$2f(0) + f'(0)$$ is equal to

$$f(0) = \begin{vmatrix}0&1&1\\2&0&6\\0&4&-2\end{vmatrix} = 0(0-24)-1(-4-0)+1(8-0) = 4+8 = 12$$.

For $$f'(0)$$, differentiate the determinant by differentiating each row separately. At $$x=0$$:

Row derivatives: $$(0, 0, 3)$$, $$(0, 2, 0)$$, $$(-1, 0, 0)$$.

$$f'(0) = \begin{vmatrix}0&0&3\\2&0&6\\0&4&-2\end{vmatrix} + \begin{vmatrix}0&1&1\\0&2&0\\0&4&-2\end{vmatrix} + \begin{vmatrix}0&1&1\\2&0&6\\-1&0&0\end{vmatrix}$$

First: $$0 - 0 + 3(8) = 24$$. Second: $$0 - 1(0) + 1(0) = 0$$. Third: $$0 - 1(0+6) + 1(0+0) = -6$$.

$$f'(0) = 24 + 0 - 6 = 18$$.

$$2f(0)+f'(0) = 24+18 = 42$$.

The correct answer is Option 3: $$42$$.