Given the constant matrix \begin{bmatrix}-3&4&-2\\1&0&1\\6&-6&5\end{bmatrix} find the fundamental matrix.
After, finding the corresponding eigenvalues which are $\lambda$ = 2,1,&-1, when trying to find the eigenvector for $\lambda$=2, I end up with the matrix
\begin{bmatrix}-5&4&-2\\1&-2&1\\6&-6&3\end{bmatrix}, then after trying to find the corresponding eigenvector I end up with u=(0,1,2), which I obtained from the reduced matrix
\begin{bmatrix} 1&0&0\\0&1&-.5\\0&0&0\end{bmatrix}
what does the final matrix imply? Does it mean that x=0, y=free parameter and z=2y, i.e. if y=1, then x=0 and z=2y?
yes, the matrix $$\begin{bmatrix} 1&0&0\\0&1&-.5\\0&0&0\end{bmatrix} \times \begin{bmatrix} 0\\y\\2y\end{bmatrix}= \begin{bmatrix} 0\\0\\0\end{bmatrix} $$
That is the eigenvector has the form $(0,y, 2y)$ as you indicated.
Now you need to find other eigenvectors and find the fundamental solution.