 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 $\hat{\mathbf{x}}$,and period averaging givesFor $\lambda >0$, this is minimized by $\mathbf{n}\parallel \hat{\mathbf{x}}$, 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: $\mathbf{n}\perp \hat{\mathbf{x}}$, 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.