Let P(E1)=P1;P(E2)=P2;P(E3)=P3
P(E1∩Eˉ2∩Eˉ3)=α=P1(1−P2)(1−P3)…(1)
P(Eˉ1∩E2∩Eˉ3)=β=(1−P1)P2(1−P3)…(2)
P(Eˉ1∩Eˉ2∩E3)=γ=(1−P1)(1−P2)P3…(3)
P(Eˉ1∩Eˉ2∩Eˉ3)=p=(1−P1)(1−P2)(1−P3)…(4)
Given that, (α−2β)p=αβ
⇒[P1(1−P2)(1−P3)−2(1−P1)P2(1−P3)]p=P1P2
(1−P1)(1−P2)(1−P3)2
⇒(P1(1−P2)−2(1−P1)P2)=P1P2
⇒(P1−P1P2−2P2+2P1P2)=P1P2
⇒P1=2P2…(1)
and similarly, (β−3γ)P=2βγ
P2=3P3…(2)
So, P1=6P3⇒P3P1=6