Consider the matrix \begin{pmatrix} 1 & -3 &1 \\ 1 & 5 & -1 \\ \ 2 & 6 &0 \\ \end{pmatrix}
This has eigenvalue 2 with 3 multiplicity. However, obtained two more eigenvector (-3 1 0) and (1 0 1). How can I found the third one to form with the Jordan form with \begin{pmatrix} 2 & 0 &0 \\ 0 & 2 & 1 \\ \ 0 & 0 &2 \\ \end{pmatrix}
https://www.mathstools.com/section/main/jordan_form_3x3_triple_root#.YI_pgS_R1N0 here is doing the same thing but I dont understand..
On
Take the last vector $v_3$ to be any vector that is not an eigenvector for $\lambda=2$, then take $v_2=(A-2I)v_3$ (which must be an eigenvector since $\operatorname{im}(A-2I)\subset\ker(A-2I)$), and finally take $v_1$ to be an eigenvalue linearly independent of $v_2$. For instance take $v_3=(1,0,0)$, $v_2=(-1,1,2)$, and $v_1=(1,0,1)$. This works because $Av_1=2v_1$, $Av_2=2v_2$, and $Av_3=v_2+2v_3$, by construction.
That matrix only has 2 independent eigenvectors, not 3.
In general, there is always exactly 1 eigenvector for each Jordan-Block, no matter how big the Jordan block is.
In your case , the eigenvalue $2$ has algebraic multiplicty = 3 (multiplicity of the root of the characteristic polynomial), but geometric multiplicity =2 (dimension of eigen-space).
You already found the eigenvectors $v_1,v_2$. It remains to find a suitable $v_3$ such that \begin{align} P&=\begin{pmatrix}v_1&v_2&v_3\end{pmatrix} \\ A&= PJP^{-1} \\ \Leftrightarrow AP&=PJ \end{align} with your Jordan-matrix $J$. From the last equation you only need the third column: \begin{align} A v_3 &= \begin{pmatrix}v_1 &v_2 &v_3\end{pmatrix}\begin{pmatrix}0\\1\\2\end{pmatrix} \\ &=v_2+2v_3 \\ \Rightarrow (A-2)v_3 &= v_2 \end{align} This is a linear equation you should be able to solve for $v_3$.
Such a recursion relation like $(A-2)v_3 = v_2$ always holds if you need additional basis-vectors for a jordan-block for which you already know the eigenvector.