Using axioms of incidence to show whether two lines meet in space

Question

I have this question:

$L_1$ and $L_2$ are lines, and $P$ is a plane. $L_1$ does not lie in $P$, but meets the plane in a point $p$. $L_2$ lies in $P$, but does not contain $p$. Does $L_1$ meet $L_2$? Which of the axioms of incidence are used in solving this question?

I think $L_1$ does not meet $L_2$ because $L_1$ and $L_2$ are not coplanar ?!

The axioms are:

1. (line axiom) Through any two distinct point there is exactly one line,
2. (plane axiom) Through any $3$-non collinear points there is exactly one plane
3. (dimension axiom) Any line contains at least two distinct points, any plane contains at least two distinct lines, there are at least two distinct planes.
4. (line-plane intersection) If two distinct points of a line lie in some plane $P$ then the whole line lies on P
5. (the parallel axiom) Let $L$ be a line and $p$ be a point then there is exactly one line that passes through $p$ and is parallel to $L$.
6. (plane plane intersection) If two distinct planes meet then their intersection is a line.

Answer 1

The intersection of a line $L_1$ with a plane $P$ in $\mathbb{R}^3$ is either empty $L_1 \cap P = \emptyset$, contains one point $L_1 \cap P = \{ p \}$ or infinite many points $L_1 \cap P = L_1$.
(This reasoning is directly from Linear Algebra, it translates into the solution of a system of linear equations, one equation expressing the plane, two equations expressing the line)

If $L_1$ lies not in $P$ and has at least one point in common you must have the case with one intersection point $p$ and you know there can not be another point from $L_1$ on $L_2$ as that one has to lie in $P$ as well and we have only one in $P$.

Update: Now let us try the axioms:

4. (line-plane intersection) If two distinct points of a line lie in some plane P then the whole line lies on P

We apply $(A \Rightarrow B) \iff (\neg B \Rightarrow A)$: $$ (\exists p, q \in L: p \ne q \wedge p, q \in P ) \Rightarrow L \subset P \iff \\ \neg (L \subset P) \Rightarrow \neg (\exists p, q \in L: p \ne q \wedge p, q \in P ) \iff \\ \neg (L \subset P) \Rightarrow (\forall: p, q \in L: p = q \vee p \notin P \vee q \notin P) $$ The last means: If $L$ does not lie in $P$ then for all two distinct points of $L$ at least one of them is not in $P$.

As $L_1$ does not lie in $P$ and $p \in P$ we know from the above $q \notin P$.

This too allows at most one point of $L_1$ to be in $P$: If we had two distinct ones both from $P$, at least one of them had not be in $P$, contradiction.

Again: We know there can not be another point $q \in L_1$ on $L_2$ as that one has with $L_2$ to lie in $P$ as well and we have only one in $P$.