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Question

If the value of the determinant ∣ ∣a111b111c∣ ∣ is positive, then

A
abc<2
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B
abc>0
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C
abc>8
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D
abc=0
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Solution

The correct option is C abc>8
Δ=∣ ∣0011ab111ac1cc∣ ∣[C1C1aC3;C2C2C3]=(1a)(1c)(b1)(1ac)=1ac+acb+abc+1ac=(2+abc)(a+b+c),as Det>0i,e.abc+2>a+b+c{a+b+c>3(abc)1/3}abc+2>3(abc)1/3x33x+2>0,(Letx=(abc)13)(x1)2(x+2)>0i.e.x>2i.e.x3=abc>8

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