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Question

If a2+b2+c2abbcca0 a,b,cR then value of the determinant ∣ ∣ ∣(ab+2)2a2b211(bc+2)2b2c2c2a21(ca+2)2∣ ∣ ∣ is

A
65
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B
a2+b2+c2+3
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C
4(a2+b2+c2)
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D
0.0
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Solution

The correct option is A 65
We have,
a2+b2+c2abbcca0
12(2a2+2b2+2c22ab2bc2ca)0
12[(ab)2+(bc)2+(ca)2]0
Sum of whole squares cannot be less than zero.
12[(ab)2+(bc)2+(ca)2]=0
So a = b = c
Putting a = b =c in the given determinant, we get
So ∣ ∣401140014∣ ∣=4(16)+1(1)=65

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