= Solution
Set $I=I_1=I_2$. The third of the <Euler equations for a torque-free rigid body> gives $I_3\dot\omega_3=0$, so $\omega_3$ is constant. Define $\Omega=(I_3-I)\omega_3/I$. The first two <Euler equations for a torque-free rigid body> become
$$
\dot\omega_1=-\Omega\omega_2,\qquad\dot\omega_2=\Omega\omega_1,
$$
so $\omega_1+i\omega_2=C e^{i\Omega t}$. Since the body-frame <angular momentum> has components $(I\omega_1,I\omega_2,I_3\omega_3)$, its transverse components rotate at $\Omega$ while its axial component is constant. Substitution of the <principal moments of inertia> gives
$$
\boxed{\Omega=\frac{3a^2-h^2}{3a^2+h^2}\,\omega_3.}
$$
This is precession relative to the body axes; the <angular momentum> is constant in an inertial frame. If $h^2=3a^2$, the rate is zero even when $\omega_3\ne0$. If the transverse components vanish, the precession cone degenerates to its axis.
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