= Solution
For a unit-speed spherical curve,
$$
\dot t=-\alpha+\kappa_g(\alpha\times t):
$$
the normal component is $-\alpha$ because
$\alpha\cdot\dot t=-1$, and the tangential normal component is the definition of $\kappa_g$. If $\kappa_g$ is constant, then
$$
\begin{aligned}
\dot e
&=\frac d{ds}(\alpha\times t+\kappa_g\alpha)\\
&=\alpha\times\dot t+\kappa_g t\\
&=\kappa_g\alpha\times(\alpha\times t)+\kappa_g t=0.
\end{aligned}
$$
Moreover $e\cdot\alpha=\kappa_g$. Hence the whole curve lies in the fixed plane
$$
\boxed{e\cdot x=\kappa_g},
$$
so it is planar.
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