By the Sylow theorems, there is only one Sylow 7-subgroup P of G. So P is normal in G and hence

G/C_G(P) = N_G(P)/C_G(P) < Aut(P)

where C_G(P) is the centralizer of P and Aut(P) is the group of automorphisms of P. Since P ~ Z₇ and Aut(Z₇) ~ Z₆, we get

|G/C_G(P)| divides 6.

Now, |G| = 385 = (5)(7)(11), so (5)(7)(11)/|C_G(P)| divides 6. It follows that |C_G(P)| = (5)(7)(11) and hence C_G(P) = G. This means that P <Z(G). Hence |P| (= 7) divides |Z(G)|.