Finding the ratio of 2 sides, and expressing OP in terms of a, b, c

I got a bit confused about this problem, and its something that my country's academic curriculum does not really delve into or practice as much.

The Problem:

Let $$A,B,C$$ be three points on a plane and $$O$$ be the origin point on this plane. Put $$\vec{a}=\vec{OA}$$, and $$\vec{c}=\vec{OC}$$, $$P$$ is a point inside the triangle $$ABC$$. Suppose that the ratio of the areas of $$\triangle PAB,\triangle PBC$$ and $$\triangle PCA$$ is $$2:3:5$$ respectively.

(1) The straight line $$BP$$ intersects the side $$AC$$ at point $$Q$$. Find $$AQ:QC$$

(2) Express $$OP$$ in terms of $$\vec {a},\vec{b},\vec{c}$$.

What method would be the most time efficient in solving this?

In addition to this, is there a proper naming etiquette in geometry, I noticed in most problems, whithout specifications, that $$\angle A$$ is always at the 12 o'clock position and the scheme continues counter clockwise.

Let $$AF$$ and $$CE$$ be altitudes of $$\Delta PAB$$ and $$\Delta PBC$$ respectively.
Thus, since $$\Delta AFQ\sim\Delta CEQ$$, we obtain: $$\frac{AQ}{QC}=\frac{AF}{CE}=\frac{\frac{1}{2}BP\cdot AF}{\frac{1}{2}BP\cdot CE}=\frac{S_{\Delta PAB}}{S_{\Delta PBC}}=\frac{2}{3}.$$
Now, you can get $$S_{\Delta PQC}$$ and $$BP:PQ$$.
I got $$\vec{OP}=\frac{3}{10}\vec{a}+\frac{1}{2}\vec{b}+\frac{1}{5}\vec{c}.$$