If the system of equations $$\alpha x + y + z = 5, x + 2y + 3z = 4, x + 3y + 5z = \beta$$. Has infinitely many solutions, then the ordered pair $$(\alpha, \beta)$$ is equal to

We need to find $$(\alpha, \beta)$$ such that the system $$\alpha x + y + z = 5$$, $$x + 2y + 3z = 4$$, $$x + 3y + 5z = \beta$$ has infinitely many solutions.

First, for infinitely many solutions, the determinant of the coefficient matrix must be zero.

$$\begin{vmatrix} \alpha & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 5 \end{vmatrix} = 0$$

$$\alpha(10 - 9) - 1(5 - 3) + 1(3 - 2) = 0$$

$$\alpha - 2 + 1 = 0 \implies \alpha = 1$$

Next, checking consistency with $$\alpha = 1$$ leads to the augmented matrix:

$$\begin{pmatrix} 1 & 1 & 1 & 5 \\ 1 & 2 & 3 & 4 \\ 1 & 3 & 5 & \beta \end{pmatrix}$$

Now performing the row operations $$R_2 - R_1$$ and $$R_3 - R_1$$ gives

$$R_2 - R_1: (0, 1, 2, -1)$$

$$R_3 - R_1: (0, 2, 4, \beta - 5)$$

Since for consistency the third row must be a multiple of the second, we require

$$(0, 2, 4, \beta - 5) = 2 \times (0, 1, 2, -1)$$

Substituting gives $$\beta - 5 = -2 \implies \beta = 3$$

Therefore, the ordered pair is $$(\alpha, \beta) = (1, 3)$$.

The answer is Option C: $$(1, 3)$$.