We know that, for infinitely many solutions,

$\Delta =0={\Delta }_x={\Delta }_y={\Delta }_z$

So, $\Delta =\left|\begin{array}{lll}\alpha & 1& 1\\ 1& 2& 3\\ 1& 3& 5\end{array}\right|=0$

$\Rightarrow \alpha \left(10-9\right)-1\left(5-3\right)+1\left(3-2\right)=0$

$\Rightarrow \alpha -2+1=0$

$\Rightarrow \alpha =1$

Now checking for ${\Delta }_x=\left|\begin{array}{lll}5& 1& 1\\ 4& 2& 3\\ \beta & 3& 5\end{array}\right|=0$

$\Rightarrow 5\left(10-9\right)-1\left(20-3\beta \right)+1\left(12-2\beta \right)$

$\Rightarrow 5-20+3\beta +12-2\beta$

$\Rightarrow -3+\beta =0$

$\Rightarrow \beta =3$

Hence $\left(\alpha ,\beta \right)=\left(1,3\right)$