My attempt:
Suppose $G$ is a group of order $84=2^2\cdot 3\cdot 7$. Then $n_7=1$, so $G$ has a unique (and therefore normal) Sylow 7-subgroup $P_7$.
Also, $n_3\in\{1,4,7,28\}$ and $n_2\in\{1,3,7,21\}$. I need to find subgroups of order $28, 21$ and $14$. This is tougher then it seemed.
$P_7$ has index $12$ in $G$. I would just say, that there is always at least one Sylow 3-subgroup $P_3$ of order 3. Now $P_3P_7\le G$ and $P_3\cap P_7 = 1$ so that $|P_3P_7|=21$. Similarly, there is a Sylow 2-subgroup $P_2$ of order 4. Then $P_2\cong C_4$ or $P_2\cong C_2\times C_2$. In the first case $P_2P_7$ is a subgroup of order 28. How do I handle the second case? I believe that this will eventually lead to a subgroup of order 14.
Thanks.
Just use the same idea. Argue that there is a subgroup $H$ of order $2$ and that $HP_7$ is a subgroup of order $14$.
Also, you don't need to consider the different possibilities for $P_2$. Again just note that $P_2P_7$ is a subgroup and that it has order $28$.