Suppose that the finite group $G$ has $61$ Sylow $3$-subgroups. Then I want to prove that there exist two Sylow $3$-subgroups $P$ and $Q$ satisfying $|P: P\cap Q|=3 $.
Since $n_{3}=61$, then the order of $G$ must be $61.3^{n}.p_{1}^{a_{1}}\dots p_{k}^{a_{k}}$. But I couldn't connect with $61$ to show the existence of $P$ and $Q$ with the desired feature. What should I look at first to prove this claim?
Consider the conjugation action of a Sylow $3$-subgroup $P$ on the set of all Sylow $3$-subgroups.
There is one fixed point, namely $P$ itself, and the other orbits have length $3^k$ for some $k>0$.
Since $60$ is not divisible by $9$, the remaining orbits cannot all have length divisible by $9$, and so there must be at least one orbit of length $3$. Let $Q$ be in such an orbit. Then the stabilizer of $Q$ in $P$ is $N_P(Q) = P \cap Q$. So $|P:P \cap Q|=3$.