To get you started some rather basic (or even trivial, depends) observations, putting $\,n_p=$ number of Sylow $p-$subgroups , $\,G:=\,$ simple group of order $\,168\,$:
$$168=2^3\cdot 3\cdot7\Longrightarrow n_7=8\;,\;n_3=7,28$$
It can't be $\,n_3=4\,$ as then $\,G\,$ has a subgroup of index 4, which is impossible since this would mean $\,G\,$ is isomorphic with a subgroup of $\,S_4\,$ . For the same reason, it must be $\,n_2=7,\,21\,$...
Perhaps reading here you'll have have the whole view. Don't worry about the number of pages as the first ones are basic results listed.
(I) If $P_i, 1 \leq i \leq 2^4$ denote the distinct 5-Sylow subgroup of $G,$ then the number of non-identity element in $\cup_{1 \leq i \leq 2^4}P_i$ is $ 2^4 \cdot 24 = 384.$ The rest of the non-identity elements must belong to 2-Sylow subgroups and a 2-Sylow subgroup contains 15 non-identity elements. So it must be unique.
(II) $P \cap Q$ is a subgroup of $P$ and $Q.$ Since $|P| = |Q| = 25,$ the only choices for $|P \cap Q|$ is 1, 5, 25 (Lagrange's theorem). By assumption $|P \cap Q| \neq 1$ and $P, Q$ are distinct. So $|P \cap Q| = 5.$
(III) $\langle P, Q \rangle$ is a subgroup of $G$ and also contains both $P$ and $Q.$ So $|\langle P, Q \rangle|$ must divide $|G|$ and also must be divisible by $|P| = |Q|$ (Lagrange's theorem).
Best Answer
$|G|=200=2^3\cdot5^2$. If you want the easy way out, Burnside's Theorem trivially proves that this is not a simple group.
By Sylow's third theorem, the amount of $5$-subgroups $n_5$ must divide $2^3=8$. However, $n_5\equiv 1\pmod 5$ so $n_5=1$.
Since conjugation preserves the order of elements and all the elements in the $5$-subgroup are the only elements which can have order that divides $5^2$, the subgroup must be normal in $G$.