= Solution
Represent the <quadratic form> by the <symmetric matrix>
$$
H=\begin{pmatrix}2&-2&0\\-2&5&0\\0&0&-1\end{pmatrix}.
$$
The $xy$ block has <characteristic polynomial> $(2-\mu)(5-\mu)-4=(\mu-1)(\mu-6)$. Corresponding orthonormal <eigenvectors>, together with the $z$ direction, are
$$
\boxed{\widetilde e_1=\frac1{\sqrt5}(2,1,0)^T,\qquad
\widetilde e_2=\frac1{\sqrt5}(-1,2,0)^T,\qquad
\widetilde e_3=(0,0,1)^T.}
$$
The <eigenvalues> are $1,6,-1$. To eliminate the linear term, solve $H\widetilde O=-(0,3\sqrt5,0)^T$, obtaining
$$
\boxed{\widetilde O=(-\sqrt5,-\sqrt5,0)^T.}
$$
Indeed the first equation gives $x=y$, and the second then gives $3y=-3\sqrt5$. Define the new coordinates by $X=\widetilde O+\widetilde x\,\widetilde e_1+\widetilde y\,\widetilde e_2+\widetilde z\,\widetilde e_3$. The original polynomial takes the value $-15$ at the center, so its translated and orthogonally diagonalized form is
$$
\widetilde x^2+6\widetilde y^2-\widetilde z^2=15.
$$
Thus
$$
\boxed{\alpha=\frac1{15},\qquad\beta=\frac25,\qquad\gamma=-\frac1{15}.}
$$
The surface is an \b[elliptic <one-sheet hyperboloid>], centered at $\widetilde O$, with axis parallel to $\widetilde e_3$. Its waist at $\widetilde z=0$ is an ellipse with semiaxes $\sqrt{15}$ and $\sqrt{5/2}$. Every constant-$\widetilde z$ section is a nonempty ellipse, so the two signs of $\widetilde z$ belong to the same connected sheet. This is an application of <principal-axis reduction of a quadric>.
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