If P != NP, then 3-SAT is not in P
The opposite is valid.

$3SAT$ is an $NP$-complete problem, so every problem in $NP$ can be reduced to $3SAT$.

If $P \neq NP$ and $3SAT \in P$, every problem in $NP$ would be in $P$, contradicting $P \neq NP$. So if $P \neq NP$ then $3SAT \notin P$.