Given,

System of linear equations,

$7x+11y+\alpha z=13$
$5x+4y+7z=\beta$
$175x+194y+57z=361$

Now we know that, for infinite solution $D=D_1=D_2=D_3=0$

Now solving $D=0$ we get,

 $\left|\begin{array}{ccc}7& 11& \alpha \\ 5& 4& 7\\ 175& 194& 57\end{array}\right|=0$

Now applying operation $R_3\to R_3-25R_1$ we get,

$\left|\begin{array}{ccc}7& 11& \alpha \\ 5& 4& 7\\ 0& -81& 57-25\alpha \end{array}\right|=0$

$\Rightarrow 81(49-5\alpha )+(57-25\alpha )(-27)=0$

$\Rightarrow 270\alpha =-2430\Rightarrow \alpha =-9$

And now solving $D_1=0$ we get,

$\left|\begin{array}{ccc}13& 11& -9\\ \beta & 4& 7\\ 361& 194& 57\end{array}\right|=0$

$\Rightarrow 13\left(4\times 57-7\times 194\right)-\beta \left(11\times 57+9\times 194\right)+361\left(11\times 7+9\times 4\right)=0$

$\Rightarrow 13\left(-1130\right)-\beta \left(2373\right)+361\left(113\right)=0$

$\Rightarrow 13\left(-10\right)-\beta \left(21\right)+361=0$

$\Rightarrow 21\beta =231$

$\Rightarrow \beta =11$

$\therefore \alpha +\beta +2=4$