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If
$|\begin{array}{ccc} a^{2} & b^{2} & c^{2} \\ (a + λ)^{2} & (b + λ)^{2} & (c + λ)^{2} \\ (a - λ)^{2} & (b - λ)^{2} & (c - λ)^{2} \end{array}| = k λ |\begin{array}{ccc} a^{2} & b^{2} & c^{2} \\ a & b & c \\ 1 & 1 & 1 \end{array}|$
$λ \neq 0$ then k is equal to:

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4λabc

-4λabc

4λ2

-4λ2

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detailed solution

Correct option is C

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

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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:

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If
$|\begin{array}{ccc} a^{2} & b^{2} & c^{2} \\ (a + λ)^{2} & (b + λ)^{2} & (c + λ)^{2} \\ (a - λ)^{2} & (b - λ)^{2} & (c - λ)^{2} \end{array}| = k λ |\begin{array}{ccc} a^{2} & b^{2} & c^{2} \\ a & b & c \\ 1 & 1 & 1 \end{array}|$
$λ \neq 0$ then k is equal to: