How can I derive whether the methyl cation or anion is planar?
The VSEPR model is a rather simple model that actually helps in this case. It assumes that anything ‘on’ a central atom — atoms that it is bound to and lone pairs — requests almost equal amounts of space, and therefore the compound’s final geometry is dictated by the number (and type; lone pairs are deemed ‘invisible’) of substituents only.

It is important to recognise that only lone pairs (and radicals) count towards the VSEPR model; empty orbitals do not. The reasoning is that something cannot take up space if it is empty.

Applying the theory to the methyl cation should give you three substituents: the three hydrogens. These three will, according to the theory, align themselves in a trigonal planar fashion around the central carbon, giving a perfect bond angle of $120^\circ$ between them. Thus, the methyl cation is planar.

Applying it to the methyl anion should give you four substituents: the three hydrogens we had previously plus the lone pair. The lone pair takes up space, so including it into our visualisation would give us a tetrahedral geometry. However, it is also invisible (we can only ‘see’ full atoms) hence the final structure is predicted to be trigonal pyramidal with the four atoms not occupying the same plane.

Therefore, assuming you copied accurately, the answer key should have given A as the correct answer.