I am working on my scholarship exam practice (high school/pre-university level) and stuck at question (2) below.
Let $A, B, C$ be three points on a plane and $O$ be the origin point on this plane. Put $\vec{a}=\vec{OA}$, $\vec{b}=\vec{OB}$, $\vec{c}=\vec{OC}$. $P$ is a point inside the triangle $ABC$. Suppose that the ratio of the areas of $\Delta PAB$, $\Delta PBC$, $\Delta PCA$ is $2:3:5$.
(1) The straight line $BP$ intersects the side $AC$ at the point $Q$. Find $AQ:QC$.
My answer is $2:3$ and it is in accordance with the answer key. The ratio was obtained from the area ratio of $\Delta PAB$ and $\Delta PBC$ which share the same base.
$Area=\frac{1}{2}(base)(height)$
So the height of both $AQ$ and $QC$ is directly proportional to its area.
My problematic question is:
(2) Express $\vec{OP}$ in terms of $\vec{a}$, $\vec{b}$, $\vec{c}$.
I honestly do not know how I should begin here. The answer key to this question is $\frac{1}{10}(3\vec{a}+5\vec{b}+2\vec{c})$. Please help.
Note: below is the actual answer key but it is in Japanese, just for reference. I do not know how we can explain from it though.
