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Find a basis for the eigenspace of a complex eigenvalue
$$(\lambda I - A)\vec x=\vec 0$$
This is the $(\lambda I - A)$:
$$\begin{bmatrix}\lambda+3 & -1 &1\\7 & \lambda-5 & 1\\-6 & -6 & \lambda+2 \end{bmatrix}$$
The characteristic polynomial is:
$$\lambda^3-64$$
My eigenvalues are: $$\lambda_1=4,\\ \lambda_2=-2+\sqrt 3 i,\\\lambda_3=-2-\sqrt 3 i$$ I solved for the real eigenvalue by finding its eigenvector $(0,1,1)^t$,but when I try to do the same with the complex eigenvalues I get that the eigenvector has to be $(0,0,0)^t$ (using a  matrix calculator ) which is false according to wolfram because the correct eigenvectors are:
$$\begin{pmatrix}\pm\frac{i}{2\sqrt3} \\ \pm\frac{i}{2\sqrt3} \\ 1\end{pmatrix}$$

The second eigenvalue $;1+2\sqrt3,i;$, right?

"My eigenvalues are $4, 1+2\sqrt 3 ,1-2\sqrt 3 i$"

By definition eigenvectors can't be the null vector.

@PanthersFan92: Looks like you lost a $2$ in two eigenvalues - check algebra, in fact, those are complex conjugate eigenvalues, so you also lost an $i$.

@PanthersFan92 I can see all the matrix's entries are real and thus the roots of its characteristic polynomial are either real or pairs of conjugated complex non-real numbers: it is impossible you have two real roots and only one complex non-real one for a real polynomial. I haven't checked but I think you're missing $;i;$ in the second root.

I got confused. I posted one of my later calculations insted of the actual eigenvalues.

The RREF for $-2 + 2i \sqrt{3} $, should be $\left(
\begin{array}{ccc}
1 & 0 & \frac{-3 i+7 \sqrt{3}}{6 \left(7 i+\sqrt{3}\right)} \
0 & 1 & \frac{-3 i+7 \sqrt{3}}{6 \left(7 i+\sqrt{3}\right)} \
0 & 0 & 0 \
\end{array}
\right)$

How did you get to that RREF?

Not that and not that: the eigenvalues are $;-2\pm2\sqrt3,i;$ , just as Moo implied in his comment.

If it's not clear, what I meant is $-2\pm\sqrt 3i=-2\pm i\sqrt 3$

And I have two complex eigenvalues, not one.
The characteristic polynomial is $\lambda^3-64$

@PanthersFan92 That's wrong, as noted above: $$z^3=64=64e^{2k\pi i}\implies z_k=4e^{\frac{2k\pi}3}\implies z_0=4,,,\color{red}{z_1=-2+2\sqrt3,i},,,\color{red}{z_2=-2-2\sqrt3,i}$$ And yes: now you have two complex (conjugated) roots, after you edited the question. When the comments above appeared you only had one.

Substitute for example$\;\lambda=-2+2\sqrt3\,i\;$ in $\;\det(\lambda I-A)\;$ to obtain the homogeneous system (remember: the system is at most of rank two!):
$$\begin{cases}&(1+2\sqrt3\,i)x-&y&+&z=0\\{}\\&7x-&(7-2\sqrt3\,i)y&+&z=0&\end{cases}\implies(-6+2\sqrt3\,i)x+(6-2\sqrt3\,i)y=0$$$${}$$
$$\implies x=y\implies(1+2\sqrt3\,i)x-x+z=0\implies z=-2\sqrt3\,i$$
and an eigenvector for this eigenvalue is for example
$$\begin{pmatrix}1\\1\\-2\sqrt3\,i\end{pmatrix}$$

Are you sure this eigenvector is correct? Emilio gave a different answer

@PanthersFan92 Of course: my eigenvector is Emilio's multiplied by a scalar different of zero. Check it.

First note that the correct eigenvalues are:
$$
\lambda_1=4 \qquad \lambda_2=-2-2i\sqrt{3}\qquad \lambda_3=-2+2i\sqrt{3}
$$
Now , to find the  corresponding eigenvector, substitute the eigenvalue in $(\lambda I -A)x=0$ and, solving the homogeneous linear system, find the corresponding eigenspace; any vector in this eigenspace is an eigenvector.
E.g., for $\lambda_3$ we have the system:
$$
\begin{cases}
(1+2i\sqrt{3})x-y+z=0\\
7x+(2i\sqrt{3}-7)y+z=0\\
-6x-6y+2i\sqrt{3}z=0
\end{cases}
$$ 
adding the first two equation we find $x=y$ and substituting:
$$
-12x+2i\sqrt{3}z=0 \Rightarrow x=\frac{2i\sqrt{3}z}{12}=\frac{iz}{2\sqrt{3}}
$$
so the eigenspace is formed by vectors of the form 
$$
\begin{bmatrix}
\frac{iz}{2\sqrt{3}}\\\frac{iz}{2\sqrt{3}}\\z
\end{bmatrix}
$$
and one possible eigenvector is
$$
\begin{bmatrix}
\frac{i}{2\sqrt{3}}\\\frac{i}{2\sqrt{3}}\\1
\end{bmatrix}
$$
You can do the same for the other eigenvalue.
