The sum of the absolute maximum and minimum values of the function $$f(x) = |x^2 - 5x + 6| - 3x + 2$$ in the interval $$[-1, 3]$$ is equal to :

$$f(x) = \vert{}x^2 - 5x + 6\vert{} - 3x + 2, \quad x \in [-1, 3]$$

$$\vert{}x^2 - 5x + 6\vert{} = \begin{cases} x^2 - 5x + 6, & x \in [-1, 2] \cup [3, 3] \\ -(x^2 - 5x + 6), & x \in (2, 3) \end{cases}$$

For $$x \in [-1, 2]$$:

$$f(x) = x^2 - 5x + 6 - 3x + 2 = x^2 - 8x + 8$$

$$f'(x) = 2x - 8 = 0 \implies x = 4 \notin [-1, 2]$$

For $$x \in (2, 3)$$:

$$f(x) = -(x^2 - 5x + 6) - 3x + 2 = -x^2 + 2x - 4$$

$$f'(x) = -2x + 2 = 0 \implies x = 1 \notin (2, 3)$$

Evaluating boundary values and turning points at $$x = -1, 2, 3$$:

$$f(-1) = (-1)^2 - 8(-1) + 8 = 1 + 8 + 8 = 17$$

$$f(2) = 2^2 - 8(2) + 8 = 4 - 16 + 8 = -4$$

$$f(3) = 3^2 - 8(3) + 8 = 9 - 24 + 8 = -7$$

Comparing values: $$\text{Absolute Maximum } (M) = 17 \quad (\text{at } x = -1)$$

$$\text{Absolute Minimum } (m) = -7 \quad (\text{at } x = 3)$$

Sum of absolute maximum and minimum values: $$M + m = 17 + (-7) = 10$$