Discussion

Explanation:

Given: xa = yb = zc = k (say)

x = ak, y = bk, z = ck 

ax-by(a+b)(x-y) + by-cz(b+c)(y-z) + cz-ax(c+a)(z-x)

a×ak-b×bk(a+b)(ak-bk) + b×bk-c×ck(b+c)(bk-ck) + c×ck-a×ak(c+a)(ck-ak)

k(a2-b2)k(a2-b2) + k(b2-c2)k(b2-c2) + k(c2-a2)k(c2-a2)

= 1 + 1 + 1 = 3

Hence, option (b).

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