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Question

∣ ∣1+x1111+y1111+z∣ ∣=

A
xyz(1+1x+1y+1z)
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B
xyz
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C
1+1x+1y+1z
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D
1x+1y+1z
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Solution

The correct option is A xyz(1+1x+1y+1z)
Δ=xyz∣ ∣ ∣ ∣1+1x1x1x1y1+1y1y1z1z1+1z∣ ∣ ∣ ∣=xyz(1+1x+1y+1z)∣ ∣ ∣1111y1+1y1y1z1z1+1z∣ ∣ ∣,by R1R1+R2+R3=xyz(1+1x+1y+1z)∣ ∣ ∣1001y101z01∣ ∣ ∣,by C2C2C1C3C3C1=xyz(1+1x+1y+1z)1001=xyz(1+1x+1y+1z).
Trick : Put x=1, y=2 and z=3, then ∣ ∣211131114∣ ∣=2(11)1(3)+1(13)=17
Option (a) gives, 1×2×3(1+11+12+13)=17.

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