Answer
See below
Work Step by Step
1. Find eigenvalues: (A-$\lambda$I)$\vec{V}$=$\vec{0}$ $\begin{bmatrix}
3-\lambda & 3 & 4\\
3 & 3-\lambda & 0\\
4 & 0 & 3-\lambda
\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0\end{bmatrix}$
$\begin{bmatrix}
3-\lambda & 3 & 4\\
3 & 3-\lambda & 0\\
4 & 0 & 3-\lambda
\end{bmatrix}=0$
$(\lambda-3)(\lambda+2)(\lambda-8)=0$
$\lambda_1=3,\lambda_2=-2,\lambda=8$
2. Find eigenvectors:
For $\lambda=-2$ let $B=A-\lambda_1I$ $B=\begin{bmatrix}
5 & 3 & 4\\
3 & 5 & 0\\
4 & 0 & 5 \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$
Let $r$ be a free variable.
$\vec{V}=r(-5,3,4)\\
E_1=\{(-\frac{1}{\sqrt 2},\frac{3}{5\sqrt 2},\frac{4}{5\sqrt 2})\} \rightarrow dim(E_1)=1$
The eigenvectors span $\{(-\frac{1}{\sqrt 2},\frac{3}{5\sqrt 2},\frac{4}{5\sqrt 2})\} $ in $R$
For $\lambda=3$ let $B=A-\lambda_1I$
$B=\begin{bmatrix}
0 & 3 & 4\\
3 & 0 & 0\\
4 & 0 & 0 \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.
$\vec{V}=r(0,-4,3) \\ E_2=\{(0,-\frac{4}{5},\frac{3}{5})\}
\rightarrow dim(E_2)=1$
The eigenvectors span $\{(0,-\frac{4}{5},\frac{3}{5})\} $ in $R$
For $\lambda=8$ let $B=A-\lambda_1I$
$B=\begin{bmatrix}
-5 & 3 & 4\\
3 & -5 & 0\\
4 & 0 & -5 \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.
$\vec{V}=r(5,3,4) \\ E_3=\{(\frac{1}{\sqrt 2},\frac{3}{5\sqrt 2},\frac{4}{5\sqrt 2})\}
\rightarrow dim(E_3)=1$
The eigenvectors span $\{(\frac{1}{\sqrt 2},\frac{3}{5\sqrt 2},\frac{4}{5\sqrt 2})\} $ in $R$
Hence, $S=\begin{bmatrix}
-\frac{1}{\sqrt 2} & 0 & \frac{1}{\sqrt 2}\\
\frac{3}{5\sqrt 2} & -\frac{4}{5} & \frac{3}{5\sqrt 2} \\
\frac{4}{5\sqrt 2} & \frac{3}{5} & \frac{4}{5\sqrt 2}
\end{bmatrix}$
then $S^TAS =diag(-2,3,8)$
