If a2b2c2(a+λ)2(b+λ)2(c+λ)2(a−λ)2(b−λ)2(c−λ)2=kλa2b2c2abc111λ ≠ 0 then k is equal to:

Using R3→R3−R2and R2→R2−R1we getΔ=a2b2c22aλ+λ22bλ+λ22cλ+λ2−4aλ−4bλ−4cλTake – 4λ common from R3, and applying R2→R2−2λR3 we getΔ=−4λ3a2b2c2111abc=λ4λ2a2b2c2abc111∴      k=4λ2