a
Sol. $7 x+11 y+\alpha z=13$
$5 x+4 y+7 z=\beta$
$175 x+194 y+57 z=361$
$\text { (i) } \times 10+(\text { ii }) \times 21-(\text { iii) }$
$z (10 \alpha+147-57)=130+21 \beta-361$
$\therefore 10 \alpha+90=0$
$\alpha=-9$
$130-361+21 \beta=0$
$\beta=11$
$\alpha+\beta+2=4$