1. Identify the species and electron count

The given species are:

• $Cl^-$
• $Ar$
• $Ca^{2+}$

Now count electrons:

• $Cl$: $17$ electrons, so $Cl^- = 18$ electrons
• $Ar = 18$ electrons
• $Ca$: $20$ electrons, so $Ca^{2+} = 18$ electrons

So all three are isoelectronic species.

2. Key rule for isoelectronic species

For isoelectronic species, the number of electrons is the same. Therefore, the main factor deciding size is the nuclear charge.

If nuclear charge increases, the same number of electrons are pulled more strongly toward the nucleus, so the size decreases.

3. Compare nuclear charges

Atomic numbers are:

• $Cl = 17$
• $Ar = 18$
• $Ca = 20$

Thus, for the isoelectronic species:

• $Cl^-$ has nuclear charge $+17$
• $Ar$ has nuclear charge $+18$
• $Ca^{2+}$ has nuclear charge $+20$

Hence, size order is: $Cl^- > Ar > Ca^{2+}$

This shows that the size is governed by nuclear charge.

4. Check the options

• A: electron-electron interaction in the outer orbitals
  Not the main differentiating factor here, since all are isoelectronic.

• B: Principal quantum number of valence shell
  All have electrons in the same outer shell $n=3$, so this is not the cause.

• C: nuclear charge
  Yes, this is the controlling factor.

• D: azimuthal quantum number of valence shell
  Outer electrons are in the same shell/subshell pattern; this is not the deciding factor here.

5. Final answer

The size of $Cl^-$, $Ar$, and $Ca^{2+}$ is affected by: $\boxed{\text{C: nuclear charge}}$