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Question

If a,b,c are unequal and different from 1 such that the points (a3a−1,a2−3a−1)(b3b−1,b2−3b−1) and (c3c−1,c2−3c−1) are collinear, then

A
bc+ca+ab+abc=0
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B
a+b+c+=abc
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C
bc+ca+ab=abc
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D
bc+ca+ababc=3(a+b+c)
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Solution

The correct option is D bc+ca+ababc=3(a+b+c)
Given point (a3a1,a23a1),(b3b1,b23b1) & (c3c1,c23c1)

They are collinear if area of Δle=0

∣ ∣ ∣ ∣ ∣ ∣ ∣a3a1a23a11b3b1b23b11c3c1c23c11∣ ∣ ∣ ∣ ∣ ∣ ∣=0
[a,b,c1]

(a1)(b1)(c1)∣ ∣ ∣a3a23a1b3b23b1c3c23c1∣ ∣ ∣=0

R1R2;R2R3

∣ ∣ ∣a3b3a2b2abb3c3b2c2bcc3c23c1∣ ∣ ∣=0

(ab)(bc)∣ ∣ ∣a2+ab+b2a+babb2+bc+c2b+cbcc3c23c1∣ ∣ ∣=0

Dividing by (ab)(bc) on both sides & then dividing by (ac)

[ac]=∣ ∣ ∣a+b+c10b2+bc+c2b+c1c3c23c1∣ ∣ ∣=0

∣ ∣ ∣a+b+c10b2+bc+c2b+c1c3c23c1∣ ∣ ∣=0

(a+b+c)b+c1c23c11b2+bc+c21c3c1=0

abcabac+3a+b2cb2bc+3b+bc2bcc2+3cb2c+b2bc2+bc+c2=0

abc(ab+bc+ca)+3(a+b+c)=0

bc+ca+ababc=3(a+b+c)

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