Solving geometry problem, in a triangle, using vectors P is the middle of the median line from vertex A, of ABC triangle. Q is the point of intersection between lines AC and BP.


Starting you off in more detail....
With $\mathbb{a, b}$ denoting the vertices $A, B$ and $C$ being the Origin, one gets $$\mathbb m = \tfrac12\mathbb b, \quad \mathbb p = \tfrac12(\mathbb{m+a})=\tfrac12\mathbb a+\tfrac14 \mathbb b$$
Now $Q$ is located on the intersection of $\vec{BP} = t\mathbb b+(1-t)\mathbb p$ and $\vec {CA} = s\mathbb a$, so we solve to get $t = -\frac13, s = \frac23$, giving $\mathbb q = \frac23 \mathbb a$.
Actually, we have all that is needed to answer the questions by now...
