If Δ=∣∣
∣∣a5−i7+i5+ib3+i7−i3−ic∣∣
∣∣, then Δ is always
A
real
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B
imaginary
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C
0
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D
None of these
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Solution
The correct option is A real Δ=∣∣
∣∣a5−i7+i5+ib3+i7−i3−ic∣∣
∣∣ =abc−a(9−i2)+(i−5)(5c+ic−(7−i)(3+i))+(7+i)[(5+i)(3−i)−b(7−i)] =abc−10a+5ic−c−25c−5ic−(i−5)(21+4i+1)+(7+i)(16−2i)−50b =abc−10a−26c−2i+114+114+2i−50b Δ=abc−10a−26c+228−50b Hence, Δ is real.