How to prove a group of order $144$ is not simple using Normalizers of Sylow intersections.
Here's what I have tried, but I am unable to proceed further.
And if I proceed with Sylow-$2$ subgroups then I am stuck as $|H\cap K|=2 \;or \; 4\; or \;8$ so I even can't proceed like this.
It looks to me a little long and confusing. I'd go as follows: suppose $\;G\;$ is simple of order $\;144=2^4\cdot3^2\;$ , so it cannot have one single Sylow $\;p - $ subgroup for any prime $\;p\;$ dividing $\;144\;$ . From here $\;n_2\in\{3,9\}\;,\;\;n_3\in\{4,16\}\;$ .
Since $\;144\nmid5!\;$ we can eliminate all the numbers less than $\;5\;$ above , otherwise the normalizer of some Sylow subgroup would have an index that number which would imply $\;G\;$ can be embedded in a group of order $\;n!\;,\;\;n\le 5\;$ , which is impossible.
Thus $\;n_2=9\;,\;\;n_3=16\;$ . Let $\;P,Q\;$ be two Sylow $\;3 - $ subgroups with non-trivial intersection -- if all of them have trivial intersection then there are $\;8\cdot16=128\;$ elements of order a divisor of $\;9\;$ , so there are left only $\;16\;$ elements that will form a unique Sylow $\;2-$ subgroup, contradiction --, so that then $\;|K:=P\cap Q|=3\;$ . Since $\;P,Q\;$ are abelian we have at least $\;9+6=15\;$ elements in $\;N_G(K)\;$ . But also $\;9\mid |N_G(K)|\;$ , so that
$$\;|N_G(K)|\ge18\implies [G:N_G(K)]\le8\implies |N_G(K)|=18\;$$
as otherwise there'd be a subgroup with index$\;\le5\;$, which was already ruled out. But a group of order $\;18=2\cdot3^2\;$ has only one single Sylow $\;3 - $ subgroup (of order $\;9\;$ , of course), and this is a contradiction and we've finished.