If the system of equations $$\\2x + 3y - z = 5\\ x + \alpha y + 3z = -4\\ 3x - y + \beta z = 7\\$$ has infinitely many solutions, then $$13\alpha\beta$$ is equal to:

The system of equations is:

$$2x + 3y - z = 5 \quad \text{(1)}$$

$$x + \alpha y + 3z = -4 \quad \text{(2)}$$

$$3x - y + \beta z = 7 \quad \text{(3)}$$

For the system to have infinitely many solutions, the equations must be linearly dependent. This means the third equation can be expressed as a linear combination of the first two. Assume:

$$\text{Eq. (3)} = p \cdot \text{Eq. (1)} + q \cdot \text{Eq. (2)}$$

Equating coefficients for each variable and the constant term:

For $$x$$: $$3 = 2p + q \quad \text{(4)}$$

For $$y$$: $$-1 = 3p + \alpha q \quad \text{(5)}$$

For $$z$$: $$\beta = -p + 3q \quad \text{(6)}$$

For constants: $$7 = 5p - 4q \quad \text{(7)}$$

Solve equations (4) and (7) for $$p$$ and $$q$$. From equation (4):

$$q = 3 - 2p$$

Substitute into equation (7):

$$5p - 4(3 - 2p) = 7$$

$$5p - 12 + 8p = 7$$

$$13p = 19$$

$$p = \frac{19}{13}$$

Then:

$$q = 3 - 2 \left( \frac{19}{13} \right) = \frac{39}{13} - \frac{38}{13} = \frac{1}{13}$$

Substitute $$p$$ and $$q$$ into equation (5) to find $$\alpha$$:

$$3 \left( \frac{19}{13} \right) + \alpha \left( \frac{1}{13} \right) = -1$$

$$\frac{57}{13} + \frac{\alpha}{13} = -1$$

Multiply both sides by 13:

$$57 + \alpha = -13$$

$$\alpha = -70$$

Substitute $$p$$ and $$q$$ into equation (6) to find $$\beta$$:

$$\beta = -\left( \frac{19}{13} \right) + 3 \left( \frac{1}{13} \right) = \frac{-19 + 3}{13} = \frac{-16}{13}$$

Now compute $$13\alpha\beta$$:

$$13 \alpha \beta = 13 \times (-70) \times \left( \frac{-16}{13} \right) = 13 \times \frac{1120}{13} = 1120$$

Thus, $$13\alpha\beta = 1120$$.

Verification: The determinant of the coefficient matrix is zero, and the linear dependence condition is satisfied, confirming infinitely many solutions.

The value 1120 corresponds to option B.