Given a $\triangle ABC$ and $P$ dividing $AB$ internally in the ratio $2:3$ $Q$ dividing $AC$ internally in the ratio $1:2$ , with $PQ$ produced and $BC$ produced intersecting in $R$ , to find the ratio in which $R$ divides $BC$.
Just by drawing a rough sketch, it is clear that $\frac {BR}{RC}<1$.
I tried solving this using vectors as follows.
Given $\vec a , \vec b , \vec c , \vec p , \vec q , \vec r$ being the position vectors of $A,B,C,P,Q,R$ respectively,
\begin{eqnarray*} \vec{p} & = & \frac{3\vec{a}+2\vec{b}}{5}\\ \therefore\vec{b} & = & \frac{5\vec{p}-3\vec{a}}{2}\\ \vec{q} & = & \frac{2\vec{a}+\vec{c}}{3}\\ \therefore\vec{c} & = & 3\vec{q}-2\vec{a} \end{eqnarray*} \begin{eqnarray*} \text{Let }\vec{r} & = & \frac{k\vec{b}+\vec{c}}{k+1}\\ \therefore\vec{r} & = & \frac{k\left(\frac{5\vec{p}-3\vec{a}}{2}\right)+\left(3\vec{q}-2\vec{a}\right)}{k+1}\\ & = & \frac{5k\vec{p}+6\vec{q}-\left(3k+4\right)\vec{a}}{2\left(k+1\right)} \end{eqnarray*}
That's where I'm stuck. Is there a way to express $\vec p$ and $\vec q$ in terms of $\vec a$ ?
Is there another way to solve this that would be easier?
Let $RB=kz$.
Construct $QS$ parallel to $AB$ and note that $\triangle CQS \sim \triangle CAB$ and that $\triangle RPB \sim \triangle RQS$. See the following figure.
Using similarity you get: $QS=\frac{10}{3}u$ and $k=9$.
I think you can finish by yourself now.