If (a − b), (b − c), (c − a) are in G.P.,

$(a-b),(b-c)$ and $(c-a)$ are in G.P.

$\therefore(b-c)^{2}=(a-b)(c-a)$

$\Rightarrow b^{2}-2 b c+c^{2}=a c-b c+a b-a^{2}$

$\Rightarrow a^{2}+b^{2}+c^{2}=a b+b c+c a$$\quad \ldots \ldots$ (i)

Now, LHS $=(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$

$=a b+b c+c a+2 a b+2 b c+2 c a \quad[$ Using (i) $]$

$=3 a b+3 b c+3 c a$

$=3(a b+b c+c a)$

$=\mathrm{RHS}$