How do you calculate the eigenvector if $2$ of $3$ equations give $0=0$? We have the matrix: $\begin{pmatrix}
7-\lambda & 12         & -8\\ 
-2        & -3-\lambda  & 4\\ 
0         & 0          & 3-\lambda 
\end{pmatrix}$
The eigenvalue is $\lambda=3$. Calculate the eigenvector.

Insert $\lambda=3$ into the matrix:
$\begin{pmatrix}
4   & 12   & -8\\ 
-2  & -6   & 4\\ 
0   &  0   & 0 
\end{pmatrix}$
We get system of equation:
$$I: 4x+12y-8z=0 \Leftrightarrow x+3y-2z=0$$
$$II: -2x-6y+4z=0 \Leftrightarrow x+3y-2z=0$$
$I: x= -3y+2z$
Insert that in $II: -3y+2z+3y-2z=0 \Leftrightarrow 0=0$
So we only have $x=-3y+2z$, awesome.. 
How can we create an eigenvector with this?
Would it work if we just set some value for $y$ and $z$? I have no idea what to do :(
 A: When calculating the eigenvalues, you may have noticed that $\lambda = 3$ was a double root of the characteristic equation or in other words: it is an eigenvalue with algebraic multiplicity 2.
This means it is possible that you find two linearly independent eigenvectors corresponding to this eigenvalue (that would mean the geometric multiplicity is 2 as well), but this isn't necessarily the case. Your work is fine so far and you end up with:

So we only have $x=-3y+2z$, awesome.. 

Now you can freely choose $y$ and $z$ (independently), yielding eigenvectors of the form:
$$\left(\begin{array}{c} x \\ y \\ z\end{array}\right)
=\left(\begin{array}{c} -3y+2z \\ y \\ z\end{array}\right)
=y\left(\begin{array}{c} -3 \\ 1 \\ 0\end{array}\right)+z\left(\begin{array}{c} 2 \\ 0 \\ 1\end{array}\right)$$
This means that any eigenvector corresponding to the eigenvalue $\lambda = 3$ can be written as a linear combination of the vectors $(-3,1,0)$ and $(2,0,1)$. These two are linearly independent so you found 2 eigenvectors; i.e. the geometric multiplicity is 2 as well.
A: you can choose any non zero vector parallel to the plane $x+3y-2z=0$ to be an eigenvector.
This situation arises whenever the characteristic equation has a double root. in this case the characteristic equation is $$(3-\lambda)(\lambda-3)(\lambda-1)=0$$
So any two independent vectors parallel to the plane can be taken as eigenvectors for the eigenvalue $3$
