Problem :
If a point P doesnt lie on the plane $(Q)$
how to prove that there is only one
plane that includes the point $P$ and
parallel to $(Q)$ ?
I think use Linear correlation of vectors
but how I don't know ?
Calle $\vec{n}$ the vector normal of
plan $(Q)$ but how ? I complete
I already to see your solution !
We want to give a (very simple) purely affine proof, without use of coordinates.
Let $\mathbb A$ denote an affine space with vector space associated $E$. If $P\in \mathbb A$, a (affine) plane $\pi$ of $\mathbb A$ through $P$ is a subspace of $\;\mathbb A\;$ of the form
$$ \pi = P + F $$
where $F$ is a vector subspace of $\;E\;$ with $\dim(F)=2$. Two planes
$$ P+F, \quad P'+F' $$
are parallel if and only if, by definition:
$$ F \subseteq F' \text{$\;\;$ or $\;\;$} F' \subseteq F $$
and therefore iff $\;F=F'$, since $\;\dim(F)=\dim(F')$.
We conclude that a plane through $P$ parallel to a given plane $Q+F$ is necessarily of the form $P+F$, what proves the existence and uniqueness.
Note that the hypothesis $P\notin Q+F$ was not used. The result is as such even if $P$ lies on the given plane $Q+F$. In that case we actually have
$$ P+F = Q + (P-Q) + F = Q + F $$
being $P-Q \in F$.