Answer
$\alpha\leq1$ for the solution to tend to zero and $a>1$ for the solution to be unbounded.
Work Step by Step
$$y''+(3-\alpha)y'-2(\alpha-1)y=0$$Let $y-e^{\lambda{t}}$ so that $(\ln{y})'=\lambda$.
$${\lambda}^2+(3-\alpha){\lambda}-2(\alpha-1)=0$$
$$\lambda_{1,2}=\frac{\alpha-3\pm\sqrt{{(3-\alpha)}^2+8(\alpha-1)}}{2}=\frac{\alpha-3\pm(\alpha+1)}{2}=\alpha-1,\alpha-2$$
$$\therefore{y}=c_{1}e^{(\alpha-1){t}}+c_{2}e^{(\alpha-2)t}$$
As $t\rightarrow+\infty$, $e^{(\alpha-1)t},{e}^{(\alpha-2)t}\rightarrow+\infty$ unless $a\leq1$.
$\therefore\alpha\leq1$ for the solution to tend to zero and $a>1$ for the solution to be unbounded.
