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Question

∣ ∣2aa+ba+cb+a2ab+cc+ac+bc∣ ∣=

A
4(b+c)(c+a)(a+b)
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B
4(b+c)(c+a)(a+b)
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C
4(b+c)(c+a)(a+b)
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D
2(b+c)(c+a)(a+b)
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Solution

The correct option is C 4(b+c)(c+a)(a+b)
Let Δ=∣ ∣2aa+ba+cb+a2ab+cc+ac+bc∣ ∣=
Putting a+b=0
b=a
then
Δ=∣ ∣2a0a+c02acac+aca2c∣ ∣=Expanding along R1
=2a{4ac(ca)2}0+(a+c){02a(c+a)}
=2a(c+a)22a(c+a)2
=0
Hence a + b is a factor of Δ similarly b + c and c + a are the factors of Δ.
On
expansion of determinant we can see that each term of the determinant
is a homogeneous expression in a, b, c of degree 3 and also R.H.S. is a
homogeneous expression of degree 3.
Let Δ=k(a+b)(b+c)(c+a)
or
∣ ∣2aa+ba+cb+a2ab+cc+ac+bc∣ ∣=k(a+b)(b+c)(c+a)
Putting a=0,b=1,c=2, we get
∣ ∣012123234∣ ∣=k(0+1)(1+2)(2+0)
01(46)+2(3+4)=6k
24=6k
k=4
Hence ∣ ∣2aa+ba+cb+a2ab+cc+ac+bc∣ ∣=4(a+b)(b+c)(c+a)

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