The correct answer is $\boxed{\left[ {\begin{array}{*{20}{c}} 1 \ { – 2} \ 3 \end{array}} \right]}$.
An eigenvector of a matrix $A$ is a nonzero vector $v$ such that there exists a scalar $\lambda$, called the eigenvalue, such that $Av=\lambda v$.
To find the eigenvalues and eigenvectors of a matrix, we can use the following steps:
- Find the characteristic polynomial of $A$, which is the determinant of the matrix $|A-\lambda I|$.
- Solve the characteristic polynomial for $\lambda$.
- For each eigenvalue $\lambda$, find all vectors $v$ such that $Av=\lambda v$. These vectors are the eigenvectors of $A$ corresponding to the eigenvalue $\lambda$.
In this case, the characteristic polynomial of $A$ is $p(\lambda)=-\lambda^3+5\lambda^2-11\lambda+6$. We can factor this polynomial as follows:
$$p(\lambda)=(\lambda-2)^2(\lambda-3)$$
Therefore, the eigenvalues of $A$ are $2$ and $3$.
To find the eigenvectors corresponding to the eigenvalue $2$, we can solve the equation $A v = 2 v$. This gives us the system of equations:
$$\begin{align}
3v_1-2v_2+2v_3 &= 0 \
0v_1-2v_2+v_3 &= 0 \
0v_1+0v_2+3v_3 &= 0
\end{align}$$
Solving this system of equations, we find that the only solution is $v=\left[ {\begin{array}{*{20}{c}} 1 \ { – 2} \ 3 \end{array}} \right]$.
Therefore, the eigenvector of $A$ corresponding to the eigenvalue $2$ is $\boxed{\left[ {\begin{array}{*{20}{c}} 1 \ { – 2} \ 3 \end{array}} \right]}$.