If the system of equations
$\begin{array}{l}\mathrm{x}+\mathrm{y}+\mathrm{z}=6\\ 2\mathrm{x}+5\mathrm{y}+\mathrm{\alpha z}=\mathrm{\beta }\\ \mathrm{x}+2\mathrm{y}+3\mathrm{z}=14\end{array}$
has infinitely many solutions, then $\mathrm{\alpha }+\mathrm{\beta }$ is equal to: 

$\begin{array}{l}\mathrm{x}+\mathrm{y}+\mathrm{z}=6 .....\left(1\right)\\ 2\mathrm{x}+5\mathrm{y}+\mathrm{\alpha z}=\mathrm{\beta } .....\left(2\right)\\ \mathrm{x}+2\mathrm{y}+3\mathrm{z}=14 ......\left(3\right)\\ \Rightarrow \mathrm{x}+\mathrm{y}=6-\mathrm{z}\\ \mathrm{and} \mathrm{x}+2\mathrm{y}=14-3\mathrm{z}\end{array}$
On solving 
$\begin{array}{r}x=z-2 and \mathrm{y}=8-2\mathrm{z}\text{ in }\left(2\right)\\ \Rightarrow 2(\mathrm{z}-2)+5(8-2\mathrm{z})+\mathrm{\alpha z}=\mathrm{\beta }\end{array}$
$\therefore (\mathrm{\alpha }-8)\mathrm{z}=\mathrm{\beta }-36$ For having infinite solutions
$\begin{array}{l}\therefore \mathrm{\alpha }-8=0\mathrm{\&}\mathrm{\beta }-36=0\\ \Rightarrow \mathrm{\alpha }=8,\mathrm{\beta }=36\left(\right)\\ \Rightarrow \boxed{\mathrm{\alpha }+\mathrm{\beta }=44}\end{array}$