The system of equations are
$$
\begin{array}{l}
x+3 y+5 z=\alpha x \\
5 x+y+3 z=\alpha y \\
3 x+5 y+z=\alpha z \\
(1-\alpha) x+3 y+5 z=0 \\
5 x+(1-\alpha) y+3 z=0 \\
3 x+5 y+(1-\alpha) z=0
\end{array}
$$
For infinite number of solutions, we must have $\left|\begin{array}{ccc}1-\alpha & 3 & 5 \\ 5 & 1-\alpha & 3 \\ 3 & 5 & 1-\alpha\end{array}\right|=0$
$\Rightarrow \quad\left|\begin{array}{ccc}9-\alpha & 3 & 5 \\ 9-\alpha & 1-\alpha & 3 \\ 9-\alpha & 5 & 1-\alpha\end{array}\right|=0,$
$C_{1} \rightarrow C_{2}+C_{2}+C_{3}$
Taking $(9-\alpha)$ common from the first column, we get $\begin{array}{l}\Rightarrow \quad(9-\alpha)\left|\begin{array}{ccc}1 & 3 & 5 \\ 1 & 1-\alpha & 3 \\ 1 & 5 & 1-\alpha\end{array}\right|=0 \\ \Rightarrow & (9-\alpha)\left|\begin{array}{ccc}1 & 3 & 5 \\ 0 & -\alpha-2 & -2 \\ 0 & 2 & -\alpha-4\end{array} \mid=0\right. \\ \Rightarrow & (9-\alpha)\left|\begin{array}{cc}\alpha+2 & 2 \\ -2 & \alpha+4\end{array}\right|=0 \\ \Rightarrow & (9-\alpha)\left(\alpha^{2}+6 \alpha+8+4\right)=0 \\ (\alpha-9)\left(\alpha^{2}+6 \alpha+12\right)=0\end{array}$
$\Rightarrow \alpha=9$ is the only real value of $\alpha$.