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Question

If a,b,cϵR, d=a+b+c, and
Δ=∣ ∣adddbdddc∣ ∣
then Δ equals

A
a2b2c2
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B
a2+b2+c2
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C
d2
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D
abcad2+2d3bd2cd2
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Solution

The correct option is C abcad2+2d3bd2cd2
Δ=∣ ∣adddbdddc∣ ∣
Applying R2R2R3
Δ=∣ ∣add0bddcddc∣ ∣=a[c(bd)d(dc)]+d[d(dc)d(bd)]=abcad2+2d3bd2cd2

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