To solve this problem, we need to compare the ionic radii of $$Ar, K$^+,$ and Ca$^{2+}. \\ \\$All three species have the same number of electrons, specifically 18 electrons,$\\$since they are isoelectronic. The electronic configuration for each is:$\\$• Ar: $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \\$• K$^+: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \\$• Ca$^{2+}: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \\ \\$The radii of isoelectronic species depend on the nuclear charge.$\\$The greater the nuclear charge, the more strongly the electrons are pulled towards the nucleus,$\\$resulting in a smaller radius.$\\ \\$Let's compare the nuclear charges:$\\$• Ar has a nuclear charge of $+18. \\$• K$^+$ has a nuclear charge of $+19. \\$• Ca$^{2+}$ has a nuclear charge of $+20. \\ \\$Therefore, the order of increasing nuclear charge is:$\\$Ar $<$ K$^+ <$ Ca$^{2+}. \\ \\$Consequently, the order of increasing radii is the reverse:$\\$Ca$^{2+} <$ K$^+ <$ Ar$. \\ \\$Thus, the correct order of increasing radii is given by Option 2:$\\$Ca$^{2+} <$ K$^+ <$ Ar$.$