For a system of 3 linear equations in 3 variables to have infinitely many solutions, we need:

• The equations to be dependent,
• So the third equation must be a linear combination of the first two,
• And the constants must satisfy the same relation.

Let us check whether the third left-hand side can be written as a combination of the first two:

$a(7x+11y+\alpha z)+b(5x+4y+7z)=175x+194y+57z$

Comparing coefficients of $x$ and $y$:

$7a+5b=175$ $11a+4b=194$

Solve these equations.

From $7a+5b=175$ $11a+4b=194$

Multiply the first by $4$: $28a+20b=700$

Multiply the second by $5$: $55a+20b=970$

Subtract: $27a=270 \implies a=10$

Then $7(10)+5b=175$ $70+5b=175$ $5b=105 \implies b=21$

Now compare the coefficient of $z$:

$a\alpha + 7b = 57$ $10\alpha + 7(21)=57$ $10\alpha +147=57$ $10\alpha=-90$ $\alpha=-9$

For infinitely many solutions, the constant terms must also satisfy the same combination:

$a(13)+b(\beta)=361$ $10(13)+21\beta=361$ $130+21\beta=361$ $21\beta=231$ $\beta=11$

Therefore,

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

Hence the correct option is:

$\boxed{4}$

which is Option B.