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

$\left[\mathrm{AD}\right]=\left[\begin{array}{cccc}7& 11& \mathrm{\alpha }& 13\\ 5& 4& 7& \mathrm{\beta }\\ 175& 194& 57& 361\end{array}\right],{\mathrm{R}}_3\to {\mathrm{R}}_3-10{\mathrm{R}}_1-21{\mathrm{R}}_2$
$\simeq \left[\begin{array}{cccc}7& 11& \mathrm{\alpha }& 13\\ 5& 4& 7& \mathrm{\beta }\\ 0& 0& -(90+10\mathrm{\alpha })& 231-21\mathrm{\beta }\end{array}\right]$
Infinite solutions $\Rightarrow 90+10\mathrm{\alpha }=0\mathrm{and}231-21\mathrm{\beta }=0$
$$\begin{gathered} \therefore \mathrm{\alpha }=-9,\mathrm{\beta }=11 \\ \mathrm{\alpha }+\mathrm{\beta }+2=4 \end{gathered}$$