let us call $$A- 2I = B = \left(\begin{array}{ccc}-2 & -3 & -2 \\1 & 1 & 1 \\1 & 2 & 1\end{array}\right).$$ row reducing we find that $B \to \pmatrix{1&0&1\\0&1&0\\0&0&0}$ so that the null of $B$ has dimension one and $$u = \pmatrix{1\\0\\-1}, Bu= 0$$ is a basis.
you can also find that null of $B^2$ has dimension $2.$ let $v$ solve $Bv = u.$ we will find $v$ by row reducing the augmented matrix $[B|u].$ we find that $$[B|u] \to \pmatrix{1&0&1&1\\0&1&0&-1\\0&0&0&0}$$ and $$v = \pmatrix{1\\-1\\0}, Bv = u$$
now, we solve $Bw = v$ and find $$[B|v] \to \pmatrix{1&0&1&-2\\0&1&0&1\\0&0&0&0}$$ and $$w = \pmatrix{-2\\1\\0}, Bw = v$$
you can verify that $\{u, v, w\}$ is a basis and with respect to this basis $B, A = B+2I$ have the representations $$ \pmatrix{0&1&0\\0&0&1\\0&0&0}, \pmatrix{2&1&0\\0&2&1\\0&0&2}$$
Your question is similar to Example 2 in Wikipedia article "Generalized eigenvectors," on which I base my answer.
For a given eigenvalue, the number of chains equals the number of linearly-independent eigenvectors for that eigenvalue. So for your matrix A, each eigenvalue has one chain.
For eigenvalue 2, because the algebraic multiplicity is one, the chain length is one and consists of the corresponding eigenvector x₁ = [–1, 1, 0]ᵀ.
For eigenvalue 3, the algebraic multiplicity is two, but there is only one corresponding eigenvector, so you need to find one more generalized eigenvector to make chain of length two. To do that, solve the matrix equation (A – 3 I)y₂ = y₁ for y₂ where y₁ is your eigenvector [1, 0, 0]ᵀ. You get y₂ = [0, 1, 1]ᵀ.
Hence, you have one chain for each eigenvector, and a chain basis is {x₁, y₁, y₂}.
With the chain basis in that order, the first Jordan block is a one-by-one block for eigenvalue 2, and the second block is a two-by-two block for eigenvalue 3.
Best Answer
As a reminder of some theory:
Suppose $M$ is an arbitrary $2 \times 2$ matrix with two eigenvalues, $\lambda_1, \lambda_2$, and hence two corresponding eigenvectors, $v_1$ and $v_2$. Define a matrix $V$ whose columns are those two eigenvectors,
\begin{equation} V =\left( \begin{matrix} v_1 & v_2 \\ | & | \end{matrix} \right) \end{equation} where those vertical bars indicate the column.
Then the action of $M$ on $V$ is the same as the action of $M$ on the individual columns. As those columns are also eigenvectors we may write
\begin{equation} MV =\left( \begin{matrix} Mv_1 & Mv_2 \\ | & | \end{matrix} \right) = \left( \begin{matrix} \lambda_1v_1 & \lambda_2v_2 \\ | & | \end{matrix} \right) = \underbrace{\left( \begin{matrix} v_1 & v_2 \\ | & | \end{matrix} \right)}_{V} \underbrace{\left( \begin{matrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{matrix} \right)}_{D} \end{equation}
where $D$ is the diagonal matrix of eigenvalues. That is, $MV = VD$. Or
\begin{equation} M = VDV^{-1} \end{equation}
This is a change of basis, to/from a basis of eigenvectors.
In this notation, the Jordan normal form is the diagonal matrix $D$. The change of basis matrix $V$ comprises of eigenvectors.
All of this generalizes to $n \times n$ matrices. And there is the added complication that you can have Jordan blocks for eigenspaces where the geometric multiplicity is less than the algebraic multiplicity. However that last wrinkle doesn't apply to your example.
Back to your problem: the basis is an appropriate basis iff it corresponds to the eigenvalues on the diagonal, $\lambda = 5, 3, 3$. To check that, apply the matrix $A$ to each of them and confirm they are eigenvectors with those eigenvalues. (Hint: they are!)