= Solution
The particle's <angular momentum> about the force centre is $\mathbf L=m\mathbf r\times\dot{\mathbf r}$. A <central force> exerts zero <torque> about that centre, so
$$
\dot{\mathbf L}=m\dot{\mathbf r}\times\dot{\mathbf r}+\mathbf r\times\mathbf F=\mathbf0.
$$
By <conservation of angular momentum>, $\mathbf L$ is a constant vector. If it is nonzero, the identity $\mathbf r\cdot\mathbf L=0$ places every point of the orbit in the fixed plane through the centre perpendicular to $\mathbf L$:
$$
\boxed{\text{The orbit lies in the plane }\mathbf r\cdot\mathbf L=0.}
$$
If $\mathbf L=0$, position and velocity are parallel. Away from the force centre, $\dot{\mathbf e}_r=(\dot{\mathbf r}-\dot r\mathbf e_r)/r=\mathbf0$, so the motion is radial along a fixed line, which lies in any plane containing that line. Thus the zero-angular-momentum case also gives planar motion. As usual, these assertions apply while the force law supplies a regular orbit; continuation through a singular force centre is a separate issue.
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