The number of real values of $\alpha$ for which the system of equations $x+3y+5z=\alpha x$, $5x+y+3z=\alpha y$, $3x+5y+z=\alpha z$ has an infinite number of solutions is:

Rewrite the system as homogeneous equations:$(1-\alpha )x+3y+5z=0$$5x+(1-\alpha )y+3z=0$$3x+5y+(1-\alpha )z=0$ For infinite solutions, the determinant of the coefficient matrix must be zero:
$$\left|\begin{array}{ccc}1-\alpha & 3& 5\\ 5& 1-\alpha & 3\\ 3& 5& 1-\alpha \end{array}\right|=0$$
Let $k=1-\alpha$:
$$\left|\begin{array}{ccc}k& 3& 5\\ 5& k& 3\\ 3& 5& k\end{array}\right|=k(k^2-15)-3(5k-9)+5(25-3k)=0$$
$k^3-15k-15k+27+125-15k=0$$k^3-45k+152=0$.Sum of rows is constant: $k+3+5=k+8$.If $k+8=0$, then $k=-8$. Substituting $k=-8$: $(-8{)}^3-45(-8)+152=-512+360+152=0$.So $(k+8)$ is a factor. Dividing $k^3-45k+152$ by $(k+8)$ gives $k^2-8k+19=0$.For $k^2-8k+19$, the discriminant $D=64-4(19)=64-76=-12<0$.This quadratic has no real roots.Thus, only $k=-8$ is a real solution.$1-\alpha =-8⟹\alpha =9$.There is only $1$ real value of $\alpha$.