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Question

If ∣∣ ∣∣y+zxyz+xzxx+yyz∣∣ ∣∣=k(x+y+z)(x−z)2, then k =

A
2xyz
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B
1
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C
xyz
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D
x2y2z2
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Solution

The correct option is B 1
∣ ∣y+zxyz+xzxx+yyz∣ ∣=(x+y+z)∣ ∣211z+xzxx+yyz∣ ∣by R1R1+R2+R3=(x+y+z)∣ ∣111xzxxyz∣ ∣; by C1C1C2=(x+y+z).{(z2xy)(xzx2)+(xyxz)}=(x+y+z)(xz)2k=1.
Trick : Put x=1, y=2, z=3, then
∣ ∣512431323∣ ∣=5(7)1(123)+2(89)=3592=24 and (x+y+z)(xz)2=(6)(2)2=24
k=2424=1.

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