Answer
See below
Work Step by Step
1. Find eigenvalues: (A-$\lambda$I)$\vec{V}$=$\vec{0}$ $\begin{bmatrix}
-3-\lambda & 2 & 2\\
2 & -3-\lambda &2\\
2 & 2 & -3-\lambda
\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0\end{bmatrix}$
$\begin{bmatrix}
-3-\lambda & 2 & 2\\
2 & -3-\lambda &2\\
2 & 2 & -3-\lambda
\end{bmatrix}=0$
$(\lambda+5)^2(\lambda-1)=0$
$\lambda_1=\lambda_2=-5,\lambda=1$
2. Find eigenvectors:
For $\lambda=1$ let $B=A-\lambda_1I$ $B=\begin{bmatrix}
-4 & 2 & 2\\
2 & -4 & 2\\
2 & 2 & -4 \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$
Let $r$ be a free variable.
$\vec{V}=r(1,1,1)\\
E_1=\{(-\frac{1}{\sqrt 2},\frac{3}{5\sqrt 2},\frac{4}{5\sqrt 2})\} \rightarrow dim(E_1)=1$
The eigenvectors span $\{(\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},\frac{1}{\sqrt 3})\} $ in $R$
For $\lambda=-5$ let $B=A-\lambda_1I$
$B=\begin{bmatrix}
2 & 2 & 2\\
2 & 2 & 2\\
2 & 2 & 2 \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix} $
Let $r,s$ be free variables.
$\vec{V}=r(-1,1,0)+s(-1,0,1) \\ E_2=\{(-\frac{1}{\sqrt 2},\frac{1}{2},0);(-\frac{1}{\sqrt 6},-\frac{1}{\sqrt 6},\frac{2}{\sqrt 6})\}
\rightarrow dim(E_2)=2$
The eigenvectors span $\{(-\frac{1}{\sqrt 2},\frac{1}{2},0);(-\frac{1}{\sqrt 6},-\frac{1}{\sqrt 6},\frac{2}{\sqrt 6})\} $ in $R$
Hence, $S=\begin{bmatrix}
\frac{1}{\sqrt 3} & -\frac{1}{\sqrt 2} & -\frac{1}{\sqrt 6}\\
\frac{1}{\sqrt 3} & \frac{1}{\sqrt 2} & \frac{-1}{\sqrt 6} \\
\frac{1}{\sqrt 3} & 0 & \frac{2}{\sqrt 6}
\end{bmatrix}$
then $S^TAS =diag(1,-5,-5)$
