Explanation
|pqrqrprpq|=0
p(rq−p2)−q(q2−rp)+r(qp−r2)=0
3pqr−p3−q3−r3=0
p3+q3+r3=3pqr
p3+q3+r3−3pqr=0
(p+q+r)(p2+q2+r2−pa−qr−pr)=0
⇒p+q+r=0 or
p2+q2+r2−pq−qr−pr=0
Given x+ay+a=0 ...(1)
bx+y+b=0 ...(2)
cx+cy+1=0 ...(3)
These lines are concurrent means they have only one intersection point
From 1 & 2,
(ab−1)y+ab−b=0
y=b−abab−1 & x=−a(b−1ab−1)
From 1 & 3,
(ac−c)y+ac−1=0
y=1−acac−c & x=−a(1−cac−c)
So,
b−abab−1=1−acac−c
⇒abc−cb+abc−a2bc=ab−a2bc−1+ac
⇒1+2abc=ab+ac+bc ...(4)
Now,
aa−1+bb−1+cc−1=ab−a+ab−bab−a−b+1+c(c−1)
=2abc−ac−bc−2ab+a+b+abc−ac−bc+cabc−ac−bc+c−ab+a+b−1
=3abc−2(ac+bc+ab)+a+b+cabc−(ac+bc+ab)+a+b+c−1
=abc−(ac+bc+ab)+(a+b+c)−1abc−(ac+bc+ab)+(a+b+c)−1=1 [From (4)]
Hence, option C.
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