= Solution
Split the mixed quadratic terms evenly across a <real symmetric matrix>:
$$
A=\begin{pmatrix}1&-2&-2\\-2&-4&2\\-2&2&1\end{pmatrix},\qquad Q_1=\mathbf r^TA\mathbf r.
$$
Its <characteristic polynomial> is
$$
\det(\lambda I-A)=(\lambda+1)(\lambda+5)(\lambda-4)=\lambda^3+2\lambda^2-19\lambda-20,
$$
so \b[$c=2$]. Choose the orthonormal <eigenvectors>
$$
u_1=\frac{(1,0,1)^T}{\sqrt2},\quad u_2=\frac{(-1,-4,1)^T}{3\sqrt2},\quad u_3=\frac{(-2,1,2)^T}{3},\qquad P=(u_1\ u_2\ u_3).
$$
The <orthogonal transformation> $\mathbf r=P\mathbf R$ gives
$$
\boxed{Q_1=-X^2-5Y^2+4Z^2,\qquad Q_2=X^2+Y^2+Z^2,\qquad\beta=-5,\quad\gamma=4.}
$$
Indeed $P^TP=I$ and $P^TAP=\operatorname{diag}(-1,-5,4)$; preserving $Q_2$ is precisely the length-preserving property of an <orthogonal matrix>.
The requested plane has normal $(2,1,-2)$. An <orthonormal basis> within it is
$$
e_1=\frac{(1,0,1)}{\sqrt2},\qquad e_2=\frac{(1,-4,-1)}{3\sqrt2}.
$$
Write a point of that plane as $\mathbf r=ue_1+we_2$. Restricting the <quadratic form> gives $Q_1=-u^2-(13/9)w^2$, so its section with $Q_1=-5$ is
$$
\boxed{\frac{u^2}{5}+\frac{w^2}{45/13}=1.}
$$
It is an <ellipse> centred at the origin, with <semi-axes> $\sqrt5$ along $e_1$ and $\sqrt{45/13}$ along $e_2$. Normalizing the in-plane basis is essential: coefficients in nonorthonormal coordinates would not directly give physical lengths.
The complete surface is the <one-sheet hyperboloid>
$$
\frac{X^2}{5}+Y^2-\frac{Z^2}{5/4}=1.
$$
It is connected, extends indefinitely in both directions along its axis $u_3$, and has an elliptical waist at $Z=0$ with <semi-axes> $\sqrt5$ and $1$. At larger $|Z|$ its transverse sections widen; its asymptotic <cone> is $4Z^2=X^2+5Y^2$. The specified plane is oblique to that axis, rather than the waist plane.
\Image[/past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2016-hyperboloid-plane-section.png]
{title=One-sheet hyperboloid in principal coordinates and its oblique elliptical section, with the intrinsic semi-axis lengths}
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