Given, $P(B)=\frac{3}{2} P(A)$ and $P(C)=\frac{1}{2} P(B)$
Since, $A, B$ and $C$ are mutually exclusive and exhaustive events.
$\begin{array}{ll}
\therefore & P(A)+P(B)+P(C)=1 \\
\Rightarrow & P(A)+\frac{3}{2} P(A)+\frac{1}{2} \times \frac{3}{2} P(A)=1 \\
\Rightarrow & P(A)\left(1+\frac{3}{2}+\frac{3}{4}\right)=1 \\
\Rightarrow & P(A)\left(\frac{13}{4}\right)=1 \Rightarrow P(A)=\frac{4}{13} \\
\therefore & P(C)=\frac{1}{2} \times \frac{3}{2} P(A)=\frac{3}{4} \times \frac{4}{13}=\frac{3}{13}
\end{array}$
Also, $A, B$ and $C$ are mutually exclusive.
$\begin{aligned}
& \therefore(A \cap B)=P(B \cap C)=P(C \cap A)=0 \\
& \text { Now, } P(A \cup C)=P(A)+P(C)-P(A \cap C) \\
& =\frac{4}{13}+\frac{3}{13}-0=\frac{7}{13}
\end{aligned}$