I'm having trouble solving this question.
What i've managed so far is. Let G be a group and $|G|=105=3*5*7$, $n_p=|Syl_p(G)|$ and let $P\in Syl_3(G)$, $Q\in Syl_5(G)$ and $R\in Syl_7(G)$. We have that
$$n_3=1 \text{ (from assumption.)}\\ n_5=1|21,\\ n_7=1|15.$$
Since the 7-Sylow and 5-Sylow groups are generated by a single element if two 5-Sylow or 7-Sylow must have a trivial intersection, otherwise they are the same group. By a counting the elements we find that at least two of the p-Sylow in G must be normal. This reduces the problem into three cases as follows.
Case 1: $n_7= 15$ and $n_5=1$:
Case 2: $n_7= 1$ and $n_5=21$:
Case 3: $n_7= 1$ and $n_5=1$:
Any suggestions on how to proceed from here?