Find an orthogonal matrix $Q$ and a diagonal matrix $D$ such that $A=QDQ^T$ 
Let $$A=\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}$$
Find an orthogonal matrix $Q$ and a diagonal matrix $D$ such that $A=QDQ^T$.

I already got these Eigenvalues $D=\begin{bmatrix}0&0&0\\0&0&0\\0&0&3\end{bmatrix}$and Eigenvectors $P=\begin{bmatrix}-1&-1&1\\1&0&1\\0&1&1\end{bmatrix}$
\begin{align}
\langle v_1,v_2\rangle &= \quad \!1+0+0=1 \\
\langle v_1,v_3\rangle &=-1+1+0=0 \\
\langle v_2,v_3\rangle &=-1+0+1=0
\end{align}
$\because \langle v_1,v_3\rangle= \langle v_2,v_3\rangle=0 \space \therefore A=A^T$
$\because \langle v_1,v_2\rangle=1+0+0=1 \space \therefore v_1\perp v_2$
But I have no idea what is the next step.
 A: I try to give my input from a methodical approach. From your above computations, we can see that:
For $\lambda = 0$, these eigenvectors $\{v_1,v_2\}$ span the eigenspace $E_{\lambda=0}$:
\begin{Bmatrix}
\begin{bmatrix}
-1\\
1 \\
0
\end{bmatrix}, 
\begin{bmatrix}
-1\\
0 \\
1
\end{bmatrix}
\end{Bmatrix}
and for $\lambda = 3$, this eigenvector $\{v_3\}$ span the eigenspace $E_{\lambda=3}$:
\begin{Bmatrix}
\begin{bmatrix}
1\\
1 \\
1
\end{bmatrix}
\end{Bmatrix}
These eigenvectors are linearly independent. However, there are repeated eigenvalues for $\lambda = 0$. Next, we check $\langle v_1, v_2 \rangle$. We observe that these eigenvectors are not orthogonal, as what you have shown $\langle v_1, v_2 \rangle =1\neq0$.
Hence, we can use Gram-Schmidt orthogonalization to obtain our orthonormal set.
For $E_{\lambda=0}$, we find our orthonormal vectors $\{e_1, e_2\}$:
\begin{align}
u_1 & = v_1 \\
e_1 & = \frac{u_1}{||u_1||} = \frac{1}{\sqrt2} 
\begin{bmatrix}
-1\\
1 \\
0
\end{bmatrix} \\ \DeclareMathOperator{\proj}{proj}
& \\
u_2 & = v_2 - \proj_{u_1} (v_2) \\
& = v_2 - \langle v_2, u_1\rangle u_1\\
& = 
\begin{bmatrix}
-1\\
0 \\
1
\end{bmatrix} -
\left\langle 
\begin{bmatrix}
-1\\
0 \\
1
\end{bmatrix}, 
\frac{1}{\sqrt2}
\begin{bmatrix}
-1\\
1 \\
0
\end{bmatrix}
\right\rangle 
\frac{1}{\sqrt2}
\begin{bmatrix}
-1\\
1 \\
0
\end{bmatrix} \\
& = 
\begin{bmatrix}
-1/2\\
-1/2 \\
1
\end{bmatrix} = \frac{1}{2}
\begin{bmatrix}
-1\\
-1 \\
2
\end{bmatrix} \\
e_2 & = \frac{u_2}{||u_2||} = \frac{1}{\sqrt6}
\begin{bmatrix}
-1 \\
-1 \\
2
\end{bmatrix} \\
\end{align}
For $E_{\lambda=3}$:
\begin{align}
u_3 & = v_3 \\
e_3 & = \frac{u_3}{||u_3||} = 
\frac{1}{\sqrt 3}
\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}
\\
\end{align}
Hence,
$$
\begin{align}
Q & = [e_1 \space e_2 \space e_3]
=
\begin{bmatrix}
-1/ \sqrt2 & -1/ \sqrt6 & 1/ \sqrt3\\
1/ \sqrt2 & -1/ \sqrt6 & 1/ \sqrt3\\
0/ \sqrt2 & 2/ \sqrt6 & 1/ \sqrt3\\
\end{bmatrix}: \\
D &
=
\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 3 \\
\end{bmatrix}
\end{align}
$$
A: Note that any linear combination of $v_1$ and $v_2$ is an eigenvector of $A$ corresponding to  the eigenvalue $0$. Moreover, as you already noted, since $A$ is a symmetric matrix, the eigenspace associated with $0$ is orthogonal to the eigenspace associated with $3$.
Therefore consider a vector $w=\alpha v_1+ \beta v_2$ with $\alpha,\beta\in \mathbb{R}$, such that $\langle v_1,w\rangle=2\alpha+\beta= 0$ and replace $v_2$ with $w$. For instance, $\alpha=1$ and $\beta=-2$ give $w=v_1 -2v_2=(1,1,-2)$.
Hence one possible orthonormal matrix $Q$ is
$$\left[\frac{v_1}{\|v_1\|} \Bigg| \frac{w}{\|w\|} \Bigg| \frac{v_3}{\|v_3\|}\right]
=\begin{bmatrix}-\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{6}}&\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{6}}&\frac{1}{\sqrt{3}}\\0&-\frac{2}{\sqrt{6}}&\frac{1}{\sqrt{3}}\end{bmatrix}
$$
A: Replace $v_2$ by
$$v'_2:=v_2-\frac{\langle v_2,v_1\rangle}{\|v_1\|^2}v_1=\begin{pmatrix}-\frac12\\-\frac12\\1\end{pmatrix}$$
(which is again in $\ker A=v_3^\perp$ since $v_1,v_2$ are, but which is now orthogonal to $v_1$).
Dividing $v_1,v'_2,v_3$ by their respective norms $\sqrt2,\sqrt\frac32,\sqrt3$ will give you the 3 columns of your matrix $Q.$
