Let $G$ be a group of order 144, and let $N_{1}$, $N_{2}$, and $N_{3}$ be of order 12 normal subgroups of $G$. Suppose that $N_{1}\cap N_{2}=N_{2}\cap N_{3}=N_{1}\cap N_{3}=\left\{e\right\}$, where $e$ is the identity of $G$. Show that $N_{1},N_{2}$,and $N_{3}$ are all abelian, and so is $G$. (Hint: Let $A$ be a group, and let both $B$ and $C$ be normal subgroups of $G$ so that $B\cap C=\left\{e\right\}$. Then, $xy=yx$ for any $x\in B$ and $y\in C$.)
It is easy to show that $G=N_{1}N_{2}=N_{2}N_{3}=N_{1}N_{3}$.
I tried to prove using the above facts and the given hint, but there is no further progress.
Give some advice! Thank you!