You get the eigenvectors just the same as you would normally, except here we need to split into cases where some of the eigenvalues may be equal.
Suppose first that none of the eigenvalues are equal. We need to find vectors $v=(a,b,c)$ such that $Av=\lambda v$. For $\lambda_1=\alpha$ for example, we have the system of equations
$$a\alpha=a\alpha$$
$$a(\beta-\alpha)+b\beta= b\alpha $$
$$a(1-\beta)+b(1-\beta)+c= c\alpha$$
The first equation is trivially true, and simplifying the second equation gives $a=\dfrac{\alpha-\beta}{\beta-\alpha}b\hspace{3mm}\implies\hspace{3mm}a=-b$, which we can include with the third equation to get $c=c\alpha\hspace{3mm}\implies\hspace{3mm}c=0$. Hence setting $a=t$, we have the eigenvectors for $\lambda_1=\alpha$ are $\left[\begin{matrix}t\\-t\\0\end{matrix}\right]$ for $t\in\mathbb{R}$.
You can apply the same process to find the eigenvectors for the other eigenvalues.
Now we have to deal with other cases: where $\alpha=\beta\ne1$, where $\alpha=1\ne\beta$, where $\beta=1\ne\alpha$, and finally where $\alpha=\beta=1$.
I will demonstrate the case $\alpha=\beta\ne1$ and leave the other cases to you.
In the $\alpha=\beta\ne1$ case, the equations for the eigenvalue $\lambda_1=\alpha$ become:
$$a\alpha=a\alpha$$
$$b\alpha= b\alpha $$
$$a(1-\alpha)+b(1-\alpha)+c= c\alpha$$
Note I just used the equations from before replaced all occurrences of $\beta$ with $\alpha$ since they are the same.
Now the first two equations are trivial and the last equation gives us $(a+b)(1-\alpha)=c(\alpha-1)\hspace{3mm}\implies\hspace{3mm} a+b=-c$. Hence setting $a=t$ and $b=s$ we get $\left[\begin{matrix}t\\s\\-(s+t)\end{matrix}\right]$ where $t,s\in\mathbb{R}$ as our eigenvectors for $\lambda_1=\lambda_2=\alpha$.
For $\lambda_3=1$, we get the system of equations
$$a\alpha=a$$
$$b\alpha= b$$
$$a(1-\alpha)+b(1-\alpha)+c= c$$
and simplification gives us $a=b=0$ and $c=c$. Hence setting $c=t$ gives us the eigenvectors for this eigenvalue as $\left[\begin{matrix}0\\0\\t\end{matrix}\right]$ for $t\in\mathbb{R}$.
I leave the rest up to you.
The searched matrix is $M=PDP^{-1}$ where $D$ is the diagonal matrix that has as diagonal elements the eigenvalues, in the same order as the eigenvectors in $P$ (see here).
$$
M=
\begin{bmatrix}
1&-1&0\\
0&1&0\\
2&0&1
\end{bmatrix}
\begin{bmatrix}
-1&0&0\\
0&0&0\\
0&0&1
\end{bmatrix}
\begin{bmatrix}
1&1&0\\
0&1&1\\
-2&-2&1
\end{bmatrix}
=
\begin{bmatrix}
-1&-1&0\\
0&0&0\\
-4&-4&1
\end{bmatrix}
$$
Best Answer
You start with the understanding of this formula: $(A-\lambda I)\vec x=0$, which is equivalent to $\det(A-\lambda I)=0$ $$\begin{vmatrix}-3-\lambda&6&-43\\0&-1-\lambda&9\\0&0&2-\lambda\end{vmatrix}=(-3-\lambda)(-1-\lambda)(2-\lambda)=0$$ Therefore, $\lambda_1=-3, \ \lambda_2=-1, \ \lambda_3=2$.
Let's do one example for eigenvectors:
Plug in the value of $\lambda$ into the augmented form of the matrix:
With $\lambda_1=-3$, $$\left[\begin{array}{ccc|c}-3-(-3)&6&-43&0\\0&-1-(-3)&9&0\\0&0&2-(-3)&0\end{array}\right]=\left[\begin{array}{ccc|c}0&6&-43&0\\0&2&9&0\\0&0&5&0\end{array}\right]$$ Solve this matrix and get $v_1=\begin{bmatrix}1\\0\\0\end{bmatrix}$
Now you can use similar approach to find the eigenvectors of the next two eigenvalues.