$(a)\;r_{Cs}+r_{Cl^{-}} =3a \qquad(b)\; r_{Cs}+r_{Cl^{-}} =\large\frac{3a}{2}\qquad(c)\;r_{Cs}+r_{Cl^{-}} = \large\frac{\sqrt{3}}{2}a\qquad(d)\;r_{Cs}+r_{Cl^{-}} =\sqrt{3}a$

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