The correct option is D Product of the common roots is abc
Assuming x2+ax+bc=0⋯(1) has roots α,β
x2+bx+ca=0⋯(2) has roots β,γ
x2+cx+ab=0⋯(3) has roots α,γ
Therefore,
α+β=−a, αβ=bc
β+γ=−b, βγ=ca
γ+α=−c, γα=ab
Adding, we get
2(α+β+γ)=−(a+b+c)⇒α+β+γ=−12(a+b+c)
Also, by multiplying, we get
α2β2γ2=a2b2c2⇒αβγ=±abc