= Solution
Let $\mathbf e_1,\mathbf e_2$ span the rotating plane and let $\mathbf n=\mathbf e_1\times\mathbf e_2$ be its normal. For modulation along $\widehat{\mathbf x}$,
$$
\mathbf p=p_0(\mathbf e_1\cos q_0x+\mathbf e_2\sin q_0x),
$$
and period averaging gives
$$
\left\langle|\nabla\mathbin\cdot\mathbf p|^2\right\rangle
=\frac{q_0^2p_0^2}{2}
[(\mathbf e_1\mathbin\cdot\widehat{\mathbf x})^2
+(\mathbf e_2\mathbin\cdot\widehat{\mathbf x})^2]
=\frac{q_0^2p_0^2}{2}[1-(\mathbf n\mathbin\cdot\widehat{\mathbf x})^2].
$$
For $\lambda>0$, this is minimized by $\mathbf n\parallel\widehat{\mathbf x}$, so the rotation plane is perpendicular to the modulation direction and the helix is transverse. For $\lambda<0$, it is minimized energetically by maximizing the bracket: $\mathbf n\perp\widehat{\mathbf x}$, so the modulation direction lies in the rotation plane. Thus either sign lifts the full rotational degeneracy, leaving only the rotations consistent with its selected relative orientation. A sufficiently small negative $\lambda$ does not overcome the stabilizing higher-gradient terms.
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