Prove that the eigenvalues of $3 \times 3$ orthogonal symmetric matrix are $\pm 1$.

Let $A$ be an orthogonal symmetric $3 \times 3$ matrix. Prove that its eigenvalues are $\pm 1$

I think you would start the proof by saying that $Av = \lambda v $ for some eigenvalue $\lambda$ and the corresponding eigenvector $v$.

then, $||Av||^2 = ||\lambda v||^2 = |\lambda|^2||v||^2$

then somehow I want to cancel the $A$ so i'm left with $||v||^2 = |\lambda|^2||v||^2$ which would then leave me with $|\lambda|^2=1$ but I'm not sure how to do the intermediate step to get there?

Hint First of all, continue your proof by noting that $\|Av\| = \|v\|$ due to the orthogonality of $A$.

Then, make the following observation. Let $\langle u,v \rangle$ denote the "dot product" of $u$ and $v$. If $Av = \lambda v$, note that the symmetry of $A$ implies that $$ \langle Av, v \rangle = \langle v, Av \rangle. $$ On the other hand, $\langle u, v \rangle = \overline{\langle v, u \rangle}$.