Actually, to each eigenvalue there belongs an eigenspace as you can easily show that if $v$ and $w$ are eigenvectors to the same eigenvalue of $A$, then $\alpha v+\beta w$ is eigenvector as well.
Therefore what we usually mean when we speak of the eigenvectors to a given eigenvalue is actually a basis for the corresponding eigenspace.
In your specific example, the eigenspace to eigenvalue $3$ is one-dimensional, therefore it has a single non-zero vector as basis. Any non-zero vector in that eigenspace will do.
Note that the eigenspace is not always one-dimensional. For example, take the identity matrix. It maps all vectors to themselves, therefore all vectors are solutions of the eigenvalue equation for eigenvalue $1$, that is, the eigenspace to eigenvalue $1$ is the full vector space. Therefore any basis of the eigenspace is a basis of the full vector space. So for the $n\times n$ identity matrix, you've got $n$ eigenvectors to the eigenvalue $1$.