Linearly independent eigenvectors The matrix A=$\begin{pmatrix}0 & 1 &1 \\1& 0&1 \\ 1 & 1 & 0\end{pmatrix}$
has eigenvalues $\lambda=2$ with algebriac multiplicity $1$ and $\lambda=-1$ with multiplicity $2$
For $\lambda=-1$
$A+I=0 \implies$$\begin{pmatrix}1 & 1 &1 \\1& 1&1 \\ 1 & 1 & 1\end{pmatrix}\begin{pmatrix}x_1  \\x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix}0  \\0 \\ 0 \end{pmatrix}$
so $x_1+x_2+x_3=0$
$3$ possible cases of eigenvectors are:
$\begin{pmatrix}-2  \\1 \\ 1 \end{pmatrix}$, $\begin{pmatrix}1  \\-2 \\ 1 \end{pmatrix}$, $\begin{pmatrix}1 \\-1 \\ 0 \end{pmatrix}$
which are all linearly independent. 
So we can have more than $2$ distinct (non-linearly dependent) eigenvectors, even if the algebraic multiplicity is only $2$?

All of the vectors you give belong to $\lambda = -1$. This set is not linearly independent; the rank of
$$
\begin{pmatrix}
 -2 & 1 & 1 \\ 1 & -2 & -1 \\ 1 & -1 & 0
\end{pmatrix}
$$
is 2, which means each two of them are sufficient to span the 2-dimensional eigenspace for this eigenvalue.
The eigenvector for $\lambda = 2$ is
$$
\begin{pmatrix}
1 \\ 1 \\ 1
\end{pmatrix}.
$$

Extension due to the question's update:
The equation $x_1+x_2+x_3=0$ has infinitely many solutions, even if you disregard scaling (i.e. impose a constraint like $x_1 = 1$). Because $\lambda = -1$ has algebraic multiplicity 2, there is no unique (up to a scalar multiple) corresponding eigenvector, but an eigenspace of dimension 2, and each vector in that eigenspace is an eigenvector. You can pick any two linearly independent vectors out of that space to construct a basis of eigenvectors (plus the one for $\lambda = 2$).
