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Question

If a,b,c are three complex numbers such that a2+b2+c2=0 and
∣ ∣ ∣(b2+c2)abacab(c2+a2)bcacbc(a2+b2)∣ ∣ ∣=K a2b2c2 then K is -

A
1
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B
2
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C
2
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D
4
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Solution

The correct option is D 4
=∣ ∣ ∣a2abacabb2bcacbcc2∣ ∣ ∣

=abc∣ ∣aaabbbccc∣ ∣

=a2b2c2∣ ∣111111111∣ ∣

=a2b2c2×[1(0)1(11)+1(1+1)]

=4a2b2c2

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