If the system of equations
$$x + y + z = 6$$
$$2x + 5y + \alpha z = \beta$$
$$x + 2y + 3z = 14$$
has infinitely many solutions, then $$\alpha + \beta$$ is equal to

We have the system: $$x + y + z = 6 \quad \cdots (1)$$ $$2x + 5y + \alpha z = \beta \quad \cdots (2)$$ $$x + 2y + 3z = 14 \quad \cdots (3)$$ and this system has infinitely many solutions, so we need to find $$\alpha + \beta$$.

For infinitely many solutions, the determinant of the coefficient matrix must be zero, and the system must be consistent. The coefficient matrix determinant is: $$D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{vmatrix}$$

Expanding along the first row: $$D = 1(15 - 2\alpha) - 1(6 - \alpha) + 1(4 - 5) = 15 - 2\alpha - 6 + \alpha - 1 = 8 - \alpha$$.

Setting $$D = 0$$: $$\alpha = 8$$.

Now for consistency with $$\alpha = 8$$, equation (2) must be a linear combination of (1) and (3). We try $$(2) = a \cdot (1) + b \cdot (3)$$: $$2 = a + b$$, $$5 = a + 2b$$, $$8 = a + 3b$$.

From the first two: $$b = 3$$, $$a = -1$$. Check: $$a + 3b = -1 + 9 = 8 = \alpha$$. Consistent.

Now $$\beta = a \cdot 6 + b \cdot 14 = -1(6) + 3(14) = -6 + 42 = 36$$.

Therefore $$\alpha + \beta = 8 + 36 = 44$$.

Hence, the correct answer is Option C: $$44$$.