If a2+b2+c2−ab−bc−ca≤0∀a,b,c∈R then value of the determinant ∣∣
∣
∣∣(a−b+2)2a2−b211(b−c+2)2b2−c2c2−a21(c−a+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
\N
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Solution
The correct option is A 65 We have, a2+b2+c2−ab−bc−ca≤0 ⇒12(2a2+2b2+2c2−2ab−2bc−2ca)≤0 ⇒12[(a−b)2+(b−c)2+(c−a)2]≤0
Sum of whole squares cannot be less than zero. ∴12[(a−b)2+(b−c)2+(c−a)2]=0
So a = b = c
Putting a = b =c in the given determinant, we get
So ∣∣
∣∣401140014∣∣
∣∣=4(16)+1(1)=65