If the system of equations x + y + z = 6 2 x + 5 y + α z = β x + 2 y + 3 z = 14 has infinitely…

$x+y+z=6 ....\left(1\right)$

$2x+5y+\alpha z=\beta ....\left(2\right)$

$x+2y+3z=14 ...\left(3\right)$

For infinite solution 

$\left|\begin{array}{ccc}1& 1& 1\\ 2& 5& \alpha \\ 1& 2& 3\end{array}\right|=0$

$\Rightarrow \left(15-2\alpha \right)-\left(6-\alpha \right)+\left(4-5\right)=0$

$\Rightarrow 8-\alpha =0\Rightarrow \alpha =8$

Also $\left|\begin{array}{ccc}1& 1& 6\\ 2& 5& \beta \\ 1& 2& 14\end{array}\right|=0$

$\Rightarrow \left(70-2\beta \right)-\left(28-\beta \right)+6\left(4-5\right)=0$

$\Rightarrow 36-\beta =0$

i.e. $\alpha =8, \beta =36$

$\Rightarrow \alpha +\beta =44$