Given a matrix $A=\begin{pmatrix} a & b & c \\ b & a & -c \\ c & -c & d \end{pmatrix}$ with positive $a,b,c,d \in R$ , which are the angle $\theta$ and rotation axis that transform $A$ into a matrix of the form $A'=\begin{pmatrix} \beta & \alpha & 0 \\ \alpha & \beta & 0 \\ 0 & 0 & \gamma \end{pmatrix}$, with $\alpha, \beta, \gamma > 0$? How can that axis and angle be written in terms of the eigenvalues of $A$?
UPDATE: Bellow you can find my solution. Since it has been obtained in a rather heuristic manner, I would like to modify the question and ask for the reasons why, given the symmetry of the matrices involved, those are the rotations and not others. Since I obtained this solution with a lack of rigor, improvements in this regard are welcome.
The rotation matrices around the axis $z$ and $x$ are
$$R_z(\theta)=\begin{pmatrix} cos\theta & -sin\theta & 0 \\ sin\theta & cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
$$R_x(\theta)=\begin{pmatrix} 1 & 0 & 0 \\ 0 & cos\theta & -sin\theta \\ 0 & sin\theta & cos\theta \end{pmatrix}$$
Let's denote the eigenvalues of $A$ as $\lambda_1 \leq \lambda_2 \leq \lambda_3$. Since $A$ is real and symmetric, the transformation that diagonalizes $A$ corresponds to a rotation. Let's define
$$ \Lambda = \begin{pmatrix} \lambda_3 & 0 & 0 \\ 0 & \lambda_2 & 0\\ 0 & 0 & \lambda_1 \end{pmatrix} $$
The rotation $R_z(\pi/4)$ transforms $\Lambda$ into a matrix with the same symmetry as $A'$,
$$ R_z(\pi/4) \Lambda R_z^T(\pi/4) = \begin{pmatrix} \beta & \alpha & 0 \\ \alpha & \beta & 0 \\ 0 & 0 & \gamma \end{pmatrix} $$
with
$$ \beta = \frac{\lambda_3+\lambda_2}{2}, \quad \alpha = \frac{\lambda_3-\lambda_2}{2}, \quad \gamma = \lambda_1$$
and therefore
$$ A' = R_z(\pi/4) \Lambda R_z^T(\pi/4)$$
On the other hand, $R_z(\pi/4)R_x(\theta)$ transforms $\Lambda$ into a matrix with the same shape as $A$,
$$ R_z(\pi/4)R_x(\theta) \Lambda R_z^T(\pi/4)R_x^T(\theta) = \begin{pmatrix} a & b & c \\ b & a & -c \\ c & -c & d \end{pmatrix} $$
with
$$ a= \frac{1}{2} \left( \lambda_3 + \lambda_1 sin^2(\theta) + \lambda_2 cos^2(\theta) \right) \\ b= \frac{1}{2} \left(\lambda_3 -\lambda_1 sin^2(\theta) - \lambda_2 cos^2(\theta) \right) \\ c = \frac{\sqrt 2}{4} (\lambda_1 - \lambda_2) sin(2\theta) \\ d = \lambda_1 cos^2(\theta) + \lambda_2 sin^2(\theta)$$
and we can therefore algebraically obtain the expressions for $a, b, c, d$ in terms of $\alpha, \beta, \gamma$ (or vice versa) since we know their relation with the eigenvalues $\lambda_1, \lambda_2, \lambda_3$. This relation depends on a single rotation angle around a rotation axis that can be obtained by diagonalizing the transformation matrix $R_{AA'}$ from $A$ to $A'$, which combining the expressions above to eliminate $\Lambda$, corresponds to
$$ R_{AA'} = R_z(\pi/4)R_x(\theta)R_z^T(\pi/4)$$