$2 x+5 y+\alpha z=\beta \quad(1)$
$x+2 y+3 z=14$
$x+y=6-z$
$x+2 y=14-3 z$
On solving
$x=z-2 \Rightarrow y=8-2 z$ in $(2)$
$2(z-2)+5(8-2 z)+\alpha z=\beta$
$(\alpha-8) z=\beta-36$ For having infinite solutions
$\alpha-8=0 \quad \& \quad \beta-36=0$
$\alpha=8, \beta=36$