= Solution
{c}
For steady <precession> at rate $\Omega$ about the fixed vertical <unit vector> $\mathbf k$,
$$
\dot{\mathbf e}=\Omega\mathbf k\wedge\mathbf e,\qquad
\ddot{\mathbf e}=\Omega^2(\cos\alpha\,\mathbf k-\mathbf e).
$$
Consequently $\mathbf e\wedge\ddot{\mathbf e}=\Omega^2\cos\alpha\,\mathbf e\wedge\mathbf k$ and $\dot{\mathbf e}=-\Omega\mathbf e\wedge\mathbf k$. Since the <axial angular velocity of a symmetric top> $n$ is constant, substitution in the equation of motion gives
$$
\left(A\Omega^2\cos\alpha-Cn\Omega\right)\mathbf e\wedge\mathbf k
=(\kappa\ell^2-mgh)\mathbf e\wedge\mathbf k.
$$
As $0<\alpha<\pi/2$, the <cross product> can be cancelled. For a specified axial spin, the required value is
$$
\boxed{\kappa=\frac{mgh+A\Omega^2\cos\alpha-Cn\Omega}{\ell^2}.}
$$
The corresponding total <angular velocity> is
$$
\boldsymbol\omega=\Omega\mathbf k+(n-\Omega\cos\alpha)\mathbf e.
$$
If the stated <angular velocity> $\Omega\mathbf k$ is the total <angular velocity> of the top, then $n=\Omega\cos\alpha$ and the requested value specializes to
$$
\boxed{\kappa=\frac{mgh+(A-C)\Omega^2\cos\alpha}{\ell^2}.}
$$
This specialization permits a direct check: $\mathbf L_Q=A\Omega\mathbf k+(C-A)\Omega\cos\alpha\,\mathbf e$, whose time <derivative> is $(A-C)\Omega^2\cos\alpha\,\mathbf e\wedge\mathbf k$, exactly the required <torque>. The first formula keeps the separate axial spin explicit when $\Omega$ is understood as the rate of <precession>.
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