If the system of linear equations
$$7x + 11y + \alpha z = 13$$
$$5x + 4y + 7z = \beta$$
$$175x + 194y + 57z = 361$$
has infinitely many solutions, then $$\alpha + \beta + 2$$ is equal to

The system of linear equations has infinitely many solutions:

$$ 7x + 11y + \alpha z = 13 \quad \text{...(1)} $$

$$ 5x + 4y + 7z = \beta \quad \text{...(2)} $$

$$ 175x + 194y + 57z = 361 \quad \text{...(3)} $$

For infinitely many solutions, equation (3) must be expressible as $$p \times (1) + q \times (2)$$.

Matching coefficients of $$x$$: $$7p + 5q = 175$$

Matching coefficients of $$y$$: $$11p + 4q = 194$$

From the first equation: $$7p + 5q = 175$$ ... (i)

From the second: $$11p + 4q = 194$$ ... (ii)

Multiply (i) by 4 and (ii) by 5:

$$28p + 20q = 700$$ and $$55p + 20q = 970$$

Subtracting: $$27p = 270 \Rightarrow p = 10$$

From (i): $$70 + 5q = 175 \Rightarrow q = 21$$

Coefficient of $$z$$: $$\alpha p + 7q = 57 \Rightarrow 10\alpha + 147 = 57 \Rightarrow \alpha = -9$$

RHS: $$13p + \beta q = 361 \Rightarrow 130 + 21\beta = 361 \Rightarrow \beta = 11$$

$$ \alpha + \beta + 2 = -9 + 11 + 2 = 4 $$

The correct answer is Option A: 4.