= Solution
Put $I=I_1=I_2$. The third of the <Euler equations for a torque-free rigid body> gives $\dot\omega_3=0$. The first two become $\dot\omega_1=-\Omega\omega_2$, $\dot\omega_2=\Omega\omega_1$, where
$$
\Omega=\frac{I_3-I}{I}\omega_3=\frac{e^2}{2-e^2}\omega_3.
$$
Therefore $\omega_1+i\omega_2=C e^{i\Omega t}$. In body coordinates, $L_1+iL_2=I C e^{i\Omega t}$ and $L_3=I_3\omega_3$ is constant. \b[The <angular momentum> precesses about the body symmetry axis], with period
$$
\boxed{P=\frac{2\pi}{|\Omega|}=\frac{2\pi(2-e^2)}{e^2|\omega_3|}.}
$$
For positive $\omega_3$ this is the printed expression. In the inertial frame the torque-free <angular momentum> is constant; the precession here is its changing representation in the rotating body frame. If $e=0$ or $\omega_3=0$, this body-frame precession rate is zero.
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