Solve this following

$\mathrm{D}=\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \alpha\end{array}\right|=\left|\begin{array}{ccc}1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 2 & \alpha-1\end{array}\right|=(\alpha-1)-4=(\alpha-5)$

for infinite solutions $\mathrm{D}=0 \Rightarrow \alpha=5$

$D_{x}=0 \Rightarrow\left|\begin{array}{lll}5 & 1 & 1 \\ 9 & 2 & 3 \\ \beta & 3 & 5\end{array}\right|=0$

$\Rightarrow\left|\begin{array}{ccc}0 & 0 & 1 \\ -1 & -1 & 3 \\ \beta-15 & -2 & 5\end{array}\right|=0$

$\Rightarrow 2+\beta-15=0 \Rightarrow \beta-13=0$

on $\beta=13$ we get $\mathrm{D}_{\mathrm{y}}=\mathrm{D}_{\mathrm{z}}=0$

$\alpha=5, \beta=13$