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I have an example:
$$A=begin{pmatrix} 2 & 2 & 4 \ 2 & 5 & 8 \ 4 & 8 & 17 end{pmatrix}$$
The eigenvalue I found is $lambda_1=lambda_2=1$ and $lambda_3=22$.

For $lambda=1$,
$$begin{pmatrix} x\ y \ z end{pmatrix}=begin{pmatrix} -2\ 1 \ 0 end{pmatrix}y+begin{pmatrix} -4\ 0 \ 1 end{pmatrix}z$$
For $lambda=22$,
$$begin{pmatrix} x\ y \ z end{pmatrix}=begin{pmatrix} 1/4\ 1/2 \ 1 end{pmatrix}z$$
However, those eigenvectors I found are not orthogonal to each other. The goal is to find an orthogonal matrix P and diagonal matrix Q so that $A=PQP^T$.

One thing we know is that eigenvectors of a symmetric matrix corresponding to distinct eigenvalues are orthogonal. So, if we find eigenvectors $v_1,v_2,v_3$ for $lambda_1< lambda_2< lambda_3$ we are done. On the other hand, we have eigenvalues $lambda_1=lambda_2=1$ and $lambda_3=22$, so that there are not $3$ distinct eigenvalues and the situation becomes somewhat more complicated.

Suppose we found $v_1,v_2in E(A,lambda_1)$ which are linearly independent (and hence a basis for the Eigenspace). We know that $v_1perp v_3$ and $v_2perp v_3$. This means $langle v_1,v_3rangle=langle v_2,v_3rangle=0$. By bilinearity of the inner product, we get that $langle av_1+bv_2,v_3rangle =0$ for all $a,bin mathbb{R}$. The upshot is that the entire eigenspace $E(A,lambda_1)$ is orthogonal to $v_3$. So, we are free to choose any basis of eigenvectors for $E(A,lambda_1)$ and proceed from there. Well, just apply Gram-Schmidt to $v_1,v_2$. Define
$$ u_1=frac{v_1}{lVert v_1rVert}$$
$$ u_2=frac{v_2-langle v_2, u_1rangle u_1}{lVert v_2-langle v_2, u_1rangle u_1rVert}.$$
A quick check shows that these two vectors form an orthonormal basis for $E(A,lambda_1)$. Then, if we take any nonzero $v_3in E(A,lambda_3)$ and set
$$ u_3=frac{v_3}{lVert v_3rVert}$$
we can see that $(u_1,u_2,u_3)$is an orthonormal eigenbasis of $mathbb{R}^3cong E(lambda_1,A)oplus E(lambda_3,A)$ with respect to $A$. You've already found the vectors $v_1,v_2,v_3$. Once you compute $u_1,u_2,u_3$, the matrix $P=[u_1,u_2,u_3]$ is orthogonal and
$$
A=P^T
begin{bmatrix}
1&0&0\
0&1&0\
0&0&22
end{bmatrix}
P.
$$

We know that the eigenvectors corresponding to different eigenvalues of a symmetric matrix are orthogonal. You have two different eigenvalues, hence you have two orthogonal eigenvectors $v_1$ and $v_2$. Since your matrix is $3times 3$, the third vector to form $P=[v_1 | v_2 |v_3]$ has to be $v_3=pm v_1times v_2$. It is easy to see that $PP^T=I$.

Now just take $Q=mathrm{diag}(lambda_1,lambda_2,lambda_3)$ and solve $A=PQP^T$ to determine $Q$ completely and then you're done.

How about Gram-Schmidt? Since the eigenspace is $2$-dimensional, there are certainly $2$ such.

Project and subtract: $(-4,0,1)-8frac15(-2,1,0)= (-frac45,-frac85,1)$.

Now normalize: $frac5{23}(-frac45,-frac85,1)=(-frac4{23},-frac8{23},frac5{23}):=b_1$. And $(-frac2{sqrt5},frac1{sqrt5},0):=b_2$.

Finally, normalize the eigenvector for $lambda =22$:
$frac{16}{21}(frac14,frac12,1)=(frac4{21},frac8{21},frac{16}{21}):=b_3$. Conveniently, this one is orthogonal to the others by symmetry of the matrix.

(Alternatively, the cross-product would have been a good way to do this as well.)

Finally, the matrix $P$ whose columns are the basis vectors, $b_1,b_2,b_3$, above will do the trick: $P^tAP=begin{pmatrix}1&0&0\0&1&0\0&0&22end{pmatrix}$.