The correct answer is $\boxed{\text{(C)}}$.
An eigenvector of a matrix $A$ is a vector $v$ such that $Av = \lambda v$ for some scalar $\lambda$. In other words, an eigenvector is a vector that is scaled by a constant when it is multiplied by the matrix.
To find the eigenvectors of a matrix, we can use the following formula:
$$v = \left[ {\begin{array}{*{20}{c}} x_1 \\ x_2 \\ x_3 \end{array}} \right]$$
$$Av = \left[ {\begin{array}{*{20}{c}} a_{11}x_1 + a_{12}x_2 + a_{13}x_3 \\ a_{21}x_1 + a_{22}x_2 + a_{23}x_3 \\ a_{31}x_1 + a_{32}x_2 + a_{33}x_3 \end{array}} \right]$$
$$\lambda v = \left[ {\begin{array}{*{20}{c}} \lambda x_1 \\ \lambda x_2 \\ \lambda x_3 \end{array}} \right]$$
$$\left[ {\begin{array}{{20}{c}} a_{11} – \lambda & a_{12} & a_{13} \\ a_{21} & a_{22} – \lambda & a_{23} \\ a_{31} & a_{32} & a_{33} – \lambda \end{array}} \right] \left[ {\begin{array}{{20}{c}} x_1 \\ x_2 \\ x_3 \end{array}} \right] = 0$$
To find the eigenvalues, we can solve the equation $|A – \lambda I| = 0$.
In this case, we have:
$$|A – \lambda I| = \left| \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 3 – \lambda \end{array} \right| = \lambda^3 – 6 \lambda^2 + 11 \lambda – 6$$
We can factor this equation as follows:
$$\lambda^3 – 6 \lambda^2 + 11 \lambda – 6 = (\lambda – 2)(\lambda – 3)(\lambda – 2)$$
Therefore, the eigenvalues of $A$ are $\lambda = 2$, $\lambda = 3$, and $\lambda = 2$.
To find the eigenvectors corresponding to each eigenvalue, we can substitute each eigenvalue into the equation $|A – \lambda I| = 0$ and solve for $x_1$, $x_2$, and $x_3$.
For $\lambda = 2$, we have:
$$\left| \begin{array}{ccc} 2 – 2 & 1 & 0 \\ 0 & 2 – 2 & 2 \\ 0 & 0 & 3 – 2 \end{array} \right| = \left| \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 1 \end{array} \right| = 0$$
Therefore, the eigenvector corresponding to $\lambda = 2$ is $v = [0 1 0]^T$.
For $\lambda = 3$, we have:
$$\left| \begin{array}{ccc} 3 – 3 & 1 & 0 \\ 0 & 3 – 3 & 2 \\ 0 & 0 & 3 – 3 \end{array} \right| = \left| \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right| = 0$$
Therefore, the eigenvector corresponding to $\lambda = 3$ is $v = [0 0 1]^T$.
For $\lambda = 2$, we have:
$$\left| \begin{array}{ccc} 2 – 2 & 1 &