Find Ratio of Area of $ARPQ$ to Area of $ABC$ In $\Delta ABC$  $P$ ,$Q$ and $R$ are points on $BC$,$CA$ and $AB$ respectively such that $BP:PC=1:2$, $CQ:QA=3:2$ and $AR:RB=3:2$
Find $Ar(ARPQ):Ar(ABC)$
Any hint please
 A: Way easier with a picture:

We have:
$$ \frac{[BPR]}{[BCA]}=\frac{BP}{BC}\cdot \frac{BR}{BA} = \frac{2}{15},$$
$$ \frac{[CPQ]}{[CAB]}=\frac{CP}{CB}\cdot\frac{CQ}{CA} = \frac{2}{5}, $$
hence:
$$ \frac{[ARPQ]}{[ABC]}=\frac{[ABC]-[BPR]-[CPQ]}{[ABC]}=1-\frac{2}{15}-\frac{2}{5}=\color{red}{\frac{7}{15}}. $$
Notice that affine maps preserve the ratios between areas, hence we may assume that $ABC$ is a right (or an equilateral) triangle without loss of generality, then use simple coordinate geometry, too.
A: Denote $S_1=Ar(PQC),  S_2=Ar(APQ), S_3=Ar(ARP), S_4=Ar(PBR), S=Ar(ABC)$.
Then you have the ratios:
$$\frac{S_1}{S_2}=\frac{CQ}{QA}=\frac{3}{2}\Rightarrow  S_1=\frac{3S_2}{2}$$
$$\frac{S_3}{S_4}=\frac{AR}{RB}=\frac{3}{2}\Rightarrow S_4=\frac{2S_3}{3}$$
$$\frac{S_1+S_2}{S_3+S_4}=\frac{CP}{PB}=\frac{2}{1}\Rightarrow \frac{3S_2/2+S_2}{S_3+2S_3/3}=\frac{2}{1}$$
which is $S_2=\frac{4S_3}{3}$
From the above equalitites we get:
$$\frac{S_2+S_3}{S}=\frac{S_2+S_3}{3S_2/2+S_2+S_3+2S_3/3}=\frac{S_2+S_3}{5S_2/2+5S_3/3}=$$
$$=\frac{4S_3/3+S_3}{10S_3/3+5S_3/3}=\frac{7}{15}$$
