In fact, eigenvectors corresponding to distinct eigenvalues are linearly independent. Let’s prove it.
Suppose that are eigenvectors corresponding to distinct eigenvalues , respectively. Let be the largest integer for which are linearly independent. If we had , we could write for some not all zero. Now , whence . As the are distinct, this implies (by linear independence) that , so , contradicting the assumption that was an eigenvector. We conclude that , so all the eigenvectors are linearly independent.
Each of the 3 eigenvalues found in part (b) has a corresponding eigenvector, and by the above, they are linearly independent. (Clearly, there cannot be more linearly independent eigenvectors, since the dimension of the matrix is 3.) Thus, the answer is