We consider the surface $S$ of the space $\mathbb{R}^3$ that is defined by the equation $2(x^2+y^2+z^2-xy-xz-yz)+3\sqrt{2}(x-z)=1$.
I want to find (using symmetric matrices) an appropriate orthonormal system of coordinates $(x_1, y_1, z_1)$ for which the above equation has the form $ax_1^2+by_1^2+cz_1^2=d$, for some $a,b,c,d\in \mathbb{R}$.
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I have done the following:
The eqquation of the surface $S$ can be written in the form $$\vec{x}^T\begin{pmatrix}2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2\end{pmatrix}\vec{x}+\begin{pmatrix}3\sqrt{2} & 0 & -3\sqrt{2}\end{pmatrix}\vec{x}=1$$ Since that matrix is symmetric it is diagonalizable with orthonormal basis.
Then we have to write the matrix $\begin{pmatrix}2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2\end{pmatrix}$ is the form $PDP^{-1}$, or not?
What do we do next?
Well, your equation is defined by a polinomial \begin{equation*} P(\vec{x} ) =\vec{x}^T A \vec{x}+v^T\vec{x}+\alpha \end{equation*} with \begin{equation*} A=\begin{pmatrix}2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2\end{pmatrix}, \quad v^T=\begin{pmatrix}3\sqrt{2} & 0 & -3\sqrt{2}\end{pmatrix}\quad \text{and} \quad \alpha=-1. \end{equation*}
Now you translate the polynomial with a vector $\vec{c} $ \begin{equation*} P(\vec{x}+\vec{c} ) =(\vec{x}+\vec{c} ) ^T A (\vec{x}+\vec{c}) +v^T(\vec{x}+\vec{c} ) +\alpha \end{equation*} \begin{equation*} P(\vec{x}+\vec{c} ) =\vec{x}^T A \vec{x}+ 2\vec{c}^T A \vec{x}+\vec{c}^T A \vec{c}+v^T \vec{x}+v^T \vec{c} +\alpha \end{equation*} \begin{equation*} P(\vec{x}+\vec{c} ) =\vec{x}^T A \vec{x}+\underline{ 2\vec{c}^T A \vec{x}+v^T \vec{x}} +\vec{c}^T A \vec{c}+v^T \vec{c} +\alpha \end{equation*}
And you want the underlined part to be zero. This gives you a new equation and solving it would give you the vector $\vec{c} $.