When applying VSEPR, imagine that the various electron pairs occupy "orbitals", in essence volumes shaped like balloons, protruding from the central atom (in this case chlorine). The electrons ("orbitals") repel each other.
Which repulsive interaction is strongest?
One assumes that, placed in equivalent relative positions (angles), LP-LP (LP=lone pair, BP=bonding pair) repulsion is strongest, followed by LP-BP and then BP-BP. The idea is that bonding electrons are more distant from the central atom, whereas nonbonding (LP) electrons are held tightly by the central atom and therefore represent a more "concentrated" charge, and therefore repel other electrons most strongly. The repulsion is also stronger the smaller the angle between the electron pairs.
The strength of repulsion determines the order in which to prioritize the interactions, but this can be a little tricky, as your example shows, because it is a "non-linear" problem. When more than one pair of LP is involved, the geometry should minimize LP-LP repulsions followed by LP-BP repulsions, but there is a tradeoff. While (c) places both LPs on opposite ends ($180^\circ$) and would seem to minimize the repulsion, there is a bit of a tradeoff between LP-LP and LP-BP repulsions. Geometry (c) results in a lot (6) of LP-BP repulsions at right angles ($90^\circ$), whereas geometry (a) has 4 LP-BP repulsions at right angles. This leads to positioning of the LPs in equatorial (!) positions, even though LP-LP distance is not maximized.
Such arguments are not transparent, they are based on a crude method that only approximates the "real" electron distribution. The experimental geometry is in fact a distorted version of (a), with the LP-BP angles slightly greater than right angles ($>90^\circ$).
My practical advice is to simply remember that according to VSEPR lone pairs are always placed in the equatorial plane in trigonal bipyramidal electron pair geometries (those with 5 pairs). See for instance the following table: