Find maximum value of $a$ such that the matrix has three linearly independent real eigenvectors . The maximum value of $a$ such that the matrix 
$\begin{pmatrix} -3 & 0 & -2 \\ 1 & -1 & 0 \\0 & a & -2\end{pmatrix}$
has three linearly independent real eigenvectors .
Please give me a hint   
Thanks
 A: EDIT: It was pointed out in the comments, that my first attempt was incorrect, since the question asks about real eigenvectors. (I only checked whether there are independent eigenvectors, admitting the possibility that they might be complex.)
I have tried to suggest an alternative solution. I hope that somebody will come up with something more simple. (Either I have made a mistake somewhere, or the computation which we have to do would be rather complicated if we want solve the question by hand.)

We can start by calculating the characteristic polynomial $\chi_A(x)$?
$\chi_A(x)=\begin{bmatrix}  x+3 & 0 & 2 \\  -1 & x+1 & 0 \\   0 & -a & x+2 \end{bmatrix} =x^3+6x^2+11x+6+2a=(x+1)(x+2)(x+3)+2a$
WolframAlpha
We can find situations for which this polynomial has real roots/distinct roots using the discriminant of the cubic polynomial.
WolframAlpha
Namely we get that for $a=\frac1{\sqrt {27}}$ there is a multiple real root. (This root can be found as the root of the derivative.) For larger $a$'s there are two complex roots and one real root. For smaller $a$'s there are three distinct real roots.
WolframAlpha
For three distinct real roots the situation is simple: How to prove that eigenvectors from different eigenvalues are linearly independent
For $a=\frac1{\sqrt {27}}$ we cannot say just from the eigenvalues whether the matrix is diagonalizable or not. So now we can try to diagonalize this matrix for this particular value of $a$. (Or at least to find the dimension of the eigenspace for the multiple root.)
WolframAlpha
WolframAlpha says that the matrix is not diagonalizable. Which means that the given matrix have three real linearly independent eigenvectors only for $a<\frac1{\sqrt {27}}$.
