A, B, C are three mutually exclusive and exhaustive events...

Given that $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are mutually exclusive and exhaustive events $\therefore \quad \mathrm{P}(\mathrm{ A})+\mathrm{P}(\mathrm{ B})+\mathrm{P}(\mathrm{C})=1$....(i) Given $3 \mathrm{P}(\mathrm{B})=4 \mathrm{P}(\mathrm{A}) \Rightarrow \mathrm{P}(\mathrm{A})=\frac{3}{4} \mathrm{P}(\mathrm{B})$ ....(ii)$and 3 \mathrm{P}(\mathrm{C})=2 \mathrm{P}(\mathrm{B}) \Rightarrow \mathrm{P}(\mathrm{C})=\frac{2}{3} \mathrm{P}(\mathrm{B})$.....(iii) Put (ii) and (iii) in (i), we get $$\begin{aligned} & \frac{3}{4} P(B)+P(B)+\frac{2}{3} P(B)=1 \\ & \Rightarrow \frac{9 P(B)+12 P(B)+8 P(B)}{12}=1 \\ & \Rightarrow \frac{29 P(B)}{12}=1 \Rightarrow P(B)=\frac{12}{29} \end{aligned}$$ Put the value of $\mathrm{P}(\mathrm{B})$ in (ii), we get $\mathrm{P}(\mathrm{ A})=\frac{3}{4} \mathrm{P}(\mathrm{ B})=\frac{3}{4} \times \frac{12}{29}=\frac{9}{29}$