= Solution
The <symmetric part> of the displayed matrix is
$$
M=\begin{pmatrix}5&0&2\\0&1&0\\2&0&2\end{pmatrix}.
$$
It equals the matrix of <8> plus $3I$, so the same <orthonormal eigenbasis> gives <eigenvalues> $6,1,1$. In coordinates $u,v,w$ along $\mathbf v_3,\mathbf v_{-2},\mathbf e_2$, respectively, the level set is
$$
6u^2+v^2+w^2=6,\qquad
\boxed{\frac{u^2}{1^2}+\frac{v^2}{(\sqrt6)^2}+\frac{w^2}{(\sqrt6)^2}=1.}
$$
All three <eigenvalues> are positive, so this is an <ellipsoid> with semi-axis lengths $1,\sqrt6,\sqrt6$. The short axis is along $(2,0,1)$, and the plane of the longest axes is $2x_1+x_3=0$.
A natural longest-axis direction in the $x_1x_3$ plane is $\mathbf v_{-2}=(1,0,-2)/\sqrt5$. The angle $\eta$ of a unit vector to the plane $x_1=0$ has $\sin\eta=|v_1|$, giving
$$
\boxed{\eta=\arcsin\frac1{\sqrt5}=\arctan\frac12\simeq26.565^\circ}
$$
for that choice. \b[Degeneracy qualification:] the two longest semi-axes are equal, so “the longest axis” does not specify a unique direction. The equally valid longest axis $\mathbf e_2$ lies in $x_1=0$ and has angle zero. Generally $\mathbf v=\cos\psi\,\mathbf v_{-2}+\sin\psi\,\mathbf e_2$ is a longest-axis unit vector, with
$$
\boxed{\eta(\psi)=\arcsin\left(\frac{|\cos\psi|}{\sqrt5}\right),\qquad
0\le\eta\le\arctan\frac12.}
$$
This is the <principal-axis freedom in a repeated ellipsoid eigenspace>; a unique answer needs an extra axis-selection convention.
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