The correct option is D None of these
Multiplying R1 by a,R2 by b and R3 by c, we have
Δ=1abc∣∣
∣
∣∣ab2c2abcab+aca2bc2abcbc+aba2b2cabcac+bc∣∣
∣
∣∣=a2b2c2abc∣∣
∣∣bc1ab+acac1bc+abab1ac+bc∣∣
∣∣=abc∣∣
∣
∣∣bc1∑abac1∑abab1∑ab∣∣
∣
∣∣{by C3→C3+C1}=abc.∑ab∣∣
∣∣bc11ca11ab11∣∣
∣∣=0, [Since C2≡C3].
Trick : Put a=1, b=2, c=3 and check it.