Hello Amanda,
I don't understand either how they got those eigenvalues, surely is a typo. Using the transformations $R_2\to R_2-R_1$, $R_3\to R_3-R_1$ and $C_1\to C_1+C_2+C_3$ we get: $$\begin{aligned}
\begin{vmatrix}{-\lambda}&{\;\;1}&{\;\;1}\\{\;\;1}&{-\lambda}&{\;\;1}\\{\;\;1}&{\;\;1}&{-\lambda}\end{vmatrix}&=\begin{vmatrix}{-\lambda}&{\;\;1}&{\;\;1}\\{\;\;1+\lambda}&{-\lambda}-1&{\;\;0}\\{\;\;1+\lambda}&{\;\;0}&{-\lambda-1}\end{vmatrix}\\&=\begin{vmatrix}{-\lambda+2}&{\;\;1}&{\;\;1}\\{\;\;0}&{-\lambda}-1&{\;\;0}\\{\;\;0}&{\;\;0}&{-\lambda-1}\end{vmatrix}\\&=(-\lambda+2)(\lambda+1)^2=0\\&\Leftrightarrow \lambda=2\mbox{ (simple) }\vee \;\lambda=-1\mbox{ (double) }
\end{aligned}$$ The eigenvectors are: $$\ker (A-2I)\equiv \left \{ \begin{matrix}-2x_1+x_2+x_3=0\\x_1-2x_2+x_3=0\\x_1+x_2-2x_3=0\end{matrix}\right. $$ As $\lambda=2$ is simple, $\dim(\ker(A-2I))=1$ and easily we find a basis of this eigenspace: $B_2=\{(1,1,1)\}$. On the other hand: $$\ker (A+I)\equiv \left \{ \begin{matrix}x_1+x_2+x_3=0\\x_1+x_2+x_3=0\\x_1+x_2+x_3=0\end{matrix}\right.$$ Now, $\dim(\ker(A+I))=3-\mbox{rank }(A+I)=3-1=2$ and easily we find a basis of this eigenspace: $B_{-1}=\{(-1,1,0),(-1,0,1)\}$.