Jordan form and an invertible $P$ for $A =\left( \begin{smallmatrix} 1&1&1 \\ 0 & 2 & 2 \\ 0 & 0 & 2 \end{smallmatrix}\right)$

You have found an eigenvector corresponding to the eigenvalue $1$, call it $v_1$, and you have one eigenvector corresponding to $2$, call it $v_2$. Suppose the last vector of the basis is $v_3$. Then, from the last column of the Jordan form we have $Av_3=v_2+2v_3$, i.e. $(A-2I)v_3=v_2$. Just solve this system to get $v_3$ (i.e. $(A-2I)x=v_2$).
Edit: Suppose $B=(v_1,v_2,v_3)$ and $P=[I]^B_E$ where $E$ is the standard basis. Then $P^{-1}AP=J$. Denote by $C_i^J$ the $i$-th column of $J$. Since $J$ is the matrix representing (the operator defined by) $A$ w.r.t basis $B$, we have $C_i^J=[Av_i]_B$. In particular, $C_3^J=[Av_3]_B=(0,1,2)^t$. Hence $Av_3=0\cdot v_1+1\cdot v_2+2\cdot v_3$.

We have $v_1=(1,0,0)^T\in \ker(A-I)$ is an eigenvector associated to the eignevalue $1$

moreover, since $\ker(1-2I)=\mathrm{span}((1,1,0)^T)$ then $v_2=(1,1,0)^T$ is an eigenvector of $A$ associted to the eigenvalue $2$ and the matrix $A$ isn't diagonalizable. Finally we look for a vector $v_3$ s.t. $Av_3=v_2+2v_3$ and we find $v_3= (0,\frac{1}{2},\frac{1}{2})^T$ hence with $$P=(v_1\ v_2\ v_3)=\left(\begin{matrix}1&1&0\\ 0&1&\frac{1}{2}\\ 0&0&\frac{1}{2}\end{matrix}\right)\quad\text{and}\quad J=\left(\begin{matrix}1&0&0\\ 0&2&1\\ 0&0&2\end{matrix}\right)$$ we have $$J=P^{-1}AP$$