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Question

The value of the determinant ∣ ∣1ab+c1bc+a1ca+b∣ ∣ is

A
a+b+c
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B
(a+b+c)2
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C
\N
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D
1+a+b+c
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Solution

The correct option is C \N
Δ=∣ ∣1ab+c1bc+a1ca+b∣ ∣=(a+b+c)∣ ∣11b+c11c+a11a+b∣ ∣(C2C2+C3)=0, (C1C2).

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