Finding the probability

Let $$E,F$$ and $$G$$ be three events such that the events $$E$$ and $$F$$ are mutually exclusive, $$P(E\cup F)=1$$, $$P(E\cap G)=1/4$$ and $$P(G)=7/12$$. Then $$P(F\cap G)=?$$

My attempt: Since $$P(E\cup F)=1$$. It means $$E\cup F$$ is the entire sample space($$S$$). So, $$E\cup F=S$$. So, $$P(E\cup F\cup G)=P(S\cup G)=P(S)+P(G)-P(S\cap G)=1+7/12-7/12=1$$ (Since the intersection of Sample space and the set G is the set G itself.)

And then using the formula for $$P(E\cup F\cup G)$$ and putting all the values. I got $$P(F\cap G)=1/3$$.

My question: Is my approach correct? Or is there any other better way to do it?

Yes, there is a slightly simpler way to solve it. Since $$E$$ and $$F$$ are mutually exclusive, and sum up to $$1$$, then $$F=\bar{E}$$. Via Law of Total Probability, we have $$P(G)=P(E\cap G)+P(\bar{E}\cap G)$$ and $$P(F\cap G)=P(\bar{E}\cap G)=7/12-1/4=1/3$$
Much simpler is to use the law of total probability. Since $$E$$ and $$F$$ is a partition of the sample space you can write directly $$P(G)=P(G\cap E) + P(G\cap F)$$.