a
$x+y+z$
$x+2 y+\alpha z=10$
$x+3 y+5 z=\beta$
$D=\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & \alpha \\ 1 & 3 & 5\end{array}\right|=1(10-3 \alpha)-1(5-\alpha)+1(3-z)$
$=10-3 \alpha-5+\alpha+1$
$=6-2 \alpha$
For unique solution $6-2 \alpha \neq 0 \Rightarrow \alpha \neq 3$