Solution: Let us write the characteristic polynomial of $A$. \begin{align*} p(t) & = \det (A - \lambda I) \\ & = \det \begin{bmatrix} -2 - t & 0 & 1 \\ -5 & 3 - t & \alpha \\ 4 & -2 & -1 - t \\ \end{bmatrix} \\ & = -(2+t) \begin{vmatrix} 3 - t & \alpha \\ -2 & -1 - t \\ \end{vmatrix} + 1 \begin{vmatrix} -5 & 3 - t \\ 4 & -2 \\ \end{vmatrix} \\ & = -(t + 2) \left( (t+1)(t - 3) + 2\alpha \right) + \left( 10 + 4 (t - 3) \right) \\ & = - (2 + t) \left( t^2 - 2t - 3 + 2\alpha \right) + 4t + 7 \\ & = -2t^2 + 4t + 6 - 4\alpha - t^3 + 2t^2 + 3t - 2t \alpha + 4t + 7 \\ & = -t^3 + (11 - 2\alpha )t + 4 - 4\alpha. \end{align*}

Since $A$ has the eigenvalues $0, 3$ and $-3$, the characteristic polynomial will be \[ p(t) = -t(t - 3)(t + 3) = -t^3 + 9t. \] Therefore, \begin{align*} & -t^3 + (11 - 2\alpha )t + 4 - 4\alpha. = -t^3 + 9t \\ \implies & t^3 + (11 - 2\alpha )t + 4 - 4\alpha = -t^3 + 9t \\ \implies & 11 - 2\alpha = 9, -4 + 4\alpha = 0 \\ \implies & \alpha = 1. \end{align*} Therefore, the required value of $\alpha $ will be $1$.