The sum of the absolute minimum and the absolute maximum values of the function
$$f(x) = |3x - x^2 + 2| - x$$ in the interval $$[-1, 2]$$ is

We need to find the sum of the absolute minimum and the absolute maximum values of $$f(x) = |3x - x^2 + 2| - x$$ on $$[-1, 2]$$. First, determine where the expression inside the absolute value changes sign. Setting $$-x^2 + 3x + 2 = 0$$ gives $$x^2 - 3x - 2 = 0$$, so $$x = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2}$$, and the relevant root in $$[-1, 2]$$ is $$r = \frac{3 - \sqrt{17}}{2} \approx -0.56$$. Since at $$x = 0$$ we have $$-0 + 0 + 2 = 2 > 0$$, it follows that $$-x^2 + 3x + 2 \geq 0$$ for $$x \in [r, 2]$$ and $$\leq 0$$ for $$x \in [-1, r]$$.

Accordingly, for $$x \in [-1, r]$$ we have $$f(x) = (x^2 - 3x - 2) - x = x^2 - 4x - 2$$, while for $$x \in [r, 2]$$ we have $$f(x) = (-x^2 + 3x + 2) - x = -x^2 + 2x + 2$$.

On the interval $$[-1, r]$$ the derivative is $$f'(x) = 2x - 4$$, which vanishes at $$x = 2$$, outside the interval, so there are no interior critical points in $$[-1, r]$$. On $$[r, 2]$$ the derivative is $$f'(x) = -2x + 2$$, which vanishes at $$x = 1$$, lying in $$[r, 2]$$.

Next, evaluate $$f(x)$$ at the boundary and critical points. At $$x = -1$$ one finds $$f(-1) = 1 + 4 - 2 = 3$$. At $$x = r$$, since $$r$$ satisfies $$r^2 - 3r - 2 = 0$$ so that $$r^2 = 3r + 2$$, we have $$f(r) = r^2 - 4r - 2 = (3r + 2) - 4r - 2 = -r = \frac{\sqrt{17} - 3}{2} \approx 0.56$$. At $$x = 1$$ we get $$f(1) = -1 + 2 + 2 = 3$$, and at $$x = 2$$ we obtain $$f(2) = -4 + 4 + 2 = 2$$.

Since the values are $$f(-1) = 3$$, $$f(r) = \frac{\sqrt{17}-3}{2}$$, $$f(1) = 3$$, and $$f(2) = 2$$, the absolute maximum is $$3$$ and the absolute minimum is $$\frac{\sqrt{17}-3}{2}$$. Therefore, their sum is $$3 + \frac{\sqrt{17} - 3}{2} = \frac{6 + \sqrt{17} - 3}{2} = \frac{\sqrt{17} + 3}{2}$$.

Thus, the correct answer is Option A: $$\frac{\sqrt{17}+3}{2}$$.