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Question

Evaluate: ∣ ∣ ∣1bca(b+c)1cab(c+a)1abc(a+b)∣ ∣ ∣

A
0
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B
1
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C
abc
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D
a2+b2+c2
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Solution

The correct option is A 0
Performing C2C2abcC3C3abc and then C3C3+C2
∣ ∣ ∣ ∣11a1a+1b+1c11b1a+1b+1c11c1a+1b+1c∣ ∣ ∣ ∣=(1a+1b+1c)∣ ∣ ∣ ∣11a111b111c1∣ ∣ ∣ ∣=0
(Two columns are identical on taking 1a+1b+1c common)

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