= Solution
Setting $\omega_1=\omega_3=0$ makes all three Euler equations stationary. The invariants then give
$$
E=\frac12I_2\omega_2^2,
\qquad
L^2=I_2^2\omega_2^2=2EI_2,
$$
so the two rotations are
$$
\omega_2=\pm\sqrt{\frac{2E}{I_2}}
=\pm\frac{L}{I_2},
$$
where $L=\sqrt{L^2}$.
Linearize about $(0,\Omega,0)$, where $\Omega$ is either of these values. Writing the transverse perturbations as $(\xi,\zeta)=(\delta\omega_1,\delta\omega_3)$ gives
$$
I_1\dot\xi=(I_2-I_3)\Omega\zeta,
\qquad
I_3\dot\zeta=(I_1-I_2)\Omega\xi.
$$
Consequently
$$
\ddot\xi
=\Omega^2\frac{(I_3-I_2)(I_2-I_1)}{I_1I_3}\,\xi.
$$
The transverse eigenvalues are therefore
$$
s=\pm |\Omega|
\sqrt{\frac{(I_3-I_2)(I_2-I_1)}{I_1I_3}},
$$
one of which is positive. Both rotations are linearly unstable, which is the <intermediate axis theorem>.
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