If the points (a3a−1,a2−3a−1),(b3b−1,b2−3b−1) and (c3c−1,c2−3c−1) are collinear for three distinct values a,b,c and a,b,c≠1 then abc−(bc+ca+ab) is

Suppose the given points lie on the line ℓx+my+n=0 General point is (t3t−1,t2−3t−1) where t=a,b,cℓ(t3t−1)+m(t2−3t−1)+n=0ℓ t3+m(t2−3)+n(t−1)=0ℓ t3+mt2+nt+(−n−3m)=0a+b+c=−mℓ, ab+bc+ca=nℓ,abc=n+3mℓabc−(bc+ca+ab)=n+3mℓ−(+nℓ)=3mℓ=3[−(a+b+c)]