$$ A = \begin{bmatrix} -3 & 3 & -2 \\ -7 & 6 & -3 \\ 1 & -1 & 2 \end{bmatrix} $$
The characteristic polynomial can be found to be $p(t)= -(t-1)(t-2)^2$. For
$t=1$, I have that $K_1 = E_1$. So $\beta_1 = \{ (1,2,1), (-1, -1,1)\}$.
$t=2$, I have that $K_2 = N(A-2I)^2$. So $\beta_2 = \{ (1, 2,0), (-1, 0, 2), (1,2,1) \} $.
Since $\beta= \beta_1 \cup \beta_2$ I have $\beta=\{ (1,2,1), (1,2,0), (-1,-1,1)\}$. Now I set
$$Q = \begin{bmatrix} -1 & -1 & 1 \\ 2 & 0 & 2 \\ 0 & 2 & 1 \end{bmatrix}$$
But then we have
$$Q^{-1}A Q = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & -1 & 2 \end{bmatrix}$$
This is close to the Jordan canonical form, but not exactly what we want. Could someone please tell me where I did something wrong?
Your kernels should be 1 and 2 dimensional respectively.
Secondly, you don't want just any basis for $\ker(A - 2I)^2$—although you can make this work through some manipulations—rather you want a generator $x$ for $\ker(A - 2I)$ and then some vector $y$ such that $(A - 2I)y = x$. Then $\{x, y\}$ is a basis for $\ker(A - 2I)^2$.
I computed the following:
\begin{align*} \ker (A - I) &= \langle (1,2,1) \rangle \\ \ker (A - 2I) &= \langle (1,1,-1) \rangle \\ (A - 2I)y = (1,1,-1) &\iff y \in (1, 2, 0) + \ker (A - 2I) \end{align*}
I am using $\langle x_1, x_2, x_3, \dots \rangle$ to denote the span.
You see that for the last step, I take a generator for $\ker (A - 2I)$ and then solve $(A - 2I)y = (1,1,-1)$ to get the second generator of $\ker (A - 2I)^2$ (rather than just computing $\ker(A - 2I)^2$ directly).
In total:
$$ Q = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 1 & 2 \\ 1 & -1 & 0 \end{pmatrix}$$