b
Let $P \left( E _{1}\right)= P _{1} ; P \left( E _{2}\right)= P _{2} ; P \left( E _{3}\right)= P _{3}$
$P \left( E _{1} \cap \overline{ E }_{2} \cap \overline{ E }_{3}\right)=\alpha= P _{1}\left(1- P _{2}\right)\left(1- P _{3}\right) \ldots \ldots$
$P \left(\overline{ E }_{1} \cap E _{2} \cap \overline{ E }_{3}\right)=\beta=\left(1- P _{1}\right) P _{2}\left(1- P _{3}\right)$
$P \left(\overline{ E }_{1} \cap \overline{ E }_{2} \cap E _{3}\right)=\gamma=\left(1- P _{1}\right)\left(1- P _{2}\right) P _{3} \ldots \ldots$
$P \left(\overline{ E }_{1} \cap \overline{ E }_{2} \cap \overline{ E }_{3}\right)= P =\left(1- P _{1}\right)\left(1- P _{2}\right)\left(1- P _{3}\right) \ldots \ldots$
Given that, $(\alpha-2 \beta) P =\alpha \beta$
$\Rightarrow\left( P _{1}\left(1- P _{2}\right)\left(1- P _{3}\right)-2\left(1- P _{1}\right) P _{2}\left(1- P _{3}\right)\right) P = P _{1} P _{2}$
$\quad\left(1- P _{1}\right)\left(1- P _{2}\right)\left(1- P _{3}\right)^{2}$
$\Rightarrow\left( P _{1}\left(1- P _{2}\right)-2\left(1- P _{1}\right) P _{2}\right)= P _{1} P _{2}$
$\Rightarrow\left( P _{1}- P _{1} P _{2}-2 P _{2}+2 P _{1} P _{2}\right)= P _{1} P _{2}$
$\Rightarrow P _{1}=2 P _{2} \quad \ldots \ldots(1)$
and similarly, $(\beta-3 \gamma) P =2 B \gamma$
$P _{2}=3 P _{3} \quad \ldots \ldots(2)$
So, $P _{1}=6 P _{3} \Rightarrow \frac{ P _{1}}{ P _{3}}=6$