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Question

If Δ=∣ ∣ ∣abcb2cc2babcc2aca2abca2bb2a∣ ∣ ∣=0 (a,b,cR
and are all different and non-zero), then the value of a+b+c is:

A
3
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B
1
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C
1
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D
0
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Solution

The correct option is D 0
Taking bc,ac and ab common from R1,R2 and R3 respectively, we have
Δ=bc.ca.ab∣ ∣abcbcacab∣ ∣=0
Expanding determinant along R1, we get
Δ=a2b2c2(a3b3c3+3abc)
a2b2c2(a3+b3+c33abc)=0
a2b2c2(a+b+c)12[(ab)2+(bc)2+(ca)2]=0
Hence we have, a+b+c=0

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