Here the way to go: consider the sequence of kernels:
The sequence stops after step $2$ since
$A-2I$ has rank $1$, hence its kernel (the eigenspace) has codimension $1$, i.e. has dimension $2$.
$(A-2I)^2$ is the null matrix, hence its kernel has dimension $3$. Take any vector in $\ker(A-2I)^2\smallsetminus\ker(A-2I)$, i.e. any vector of $\mathbf R^3$ which is not an eigenvector. As the eigenspace is defined by the equation $\; y=2x$, we'll take, say
Note $e'_2=(A-2I)e'_3=(1,2,1),\;$ is an eigenvector by construction1,2,0. We complete this set of two vectors to a basis, by choosing another eigenvector, linearly independent from $e'_2$, say
The definition of $e'_2$ from $e'_3$ can be written as $\; Ae'_3=2e'_3+e'_2$, so the matrix of the lineap map in basis $(e'_1,e'_2,e'_3)$ is the Jordan form: