= Solution
A <fixed space frame> is an inertial orthonormal frame whose axes remain fixed in space. A <principal body frame> is attached to the rigid body and aligned with the principal axes of its inertia tensor, so that the tensor is diagonal with entries $I_1,I_2,I_3$.
If $e_i(t)$ are the body axes and $\omega$ is the <angular velocity>, then the <derivative of a body-fixed basis vector> is
$$
\dot e_i=\omega\times e_i.
$$
In the principal frame,
$$
L=\sum_{i=1}^3I_i\omega_i e_i.
$$
For torque-free motion, the <angular momentum> has zero space derivative. Therefore
$$
0=\frac{dL}{dt}
=\sum_iI_i\dot\omega_i e_i+\omega\times L.
$$
Taking body-frame components gives the <Euler equations for a torque-free rigid body>:
$$
\begin{aligned}
I_1\dot\omega_1&=(I_2-I_3)\omega_2\omega_3,\\
I_2\dot\omega_2&=(I_3-I_1)\omega_3\omega_1,\\
I_3\dot\omega_3&=(I_1-I_2)\omega_1\omega_2.
\end{aligned}
$$
For an axisymmetric body, put $I_1=I_2=I$ and let $e_3$ be its symmetry axis. Decompose $\omega=\omega_1e_1+\omega_2e_2+\omega_3e_3$. Then
$$
\begin{aligned}
L
&=I\omega_1e_1+I\omega_2e_2+I_3\omega_3e_3\\
&=I\omega+(I_3-I)\omega_3e_3.
\end{aligned}
$$
\b[Thus $L$ is a linear combination of $\omega$ and $e_3$, proving that the angular momentum, angular velocity, and symmetry axis are always coplanar. This is the <coplanarity in a torque-free axisymmetric rigid body>.]
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