GIVEN :-

r = 0.1278 nm = 0.1278 x 10^-9 m.
For FCC, a = 2 (square root of 2) r
i.e. a = 2 x 1.414 x 0.1278 x 10^-9 m
i.e. a = 3.61 x 10^-10m

Spacing of (hkl) plane = d_hkl = [a / square root of ( h² + k² + l² )]

So, spacing of (111) plane = d₁₁₁ = [ 3.61 x 10^-10 / square root of ( 1² + 1² + 1² )]
i.e. d₁₁₁ = 2.08 x 10^-10m.

So, spacing of (321) plane = d₃₂₁ = [ 3.61 x 10^-10 / square root of ( 3² + 2² + 1² )]
i.e. d₃₂₁ = 9.65 x 10^-11m.