mass point geometry involving split masses The question:
Let $ABC$ be a triangle. $D$ and $E$ are points on $AB$ such that $AD:DE:EB=2:3:2$.
$F$ and $G$ are points on $BC$ such that $BF:FG:GC=1:3:2$. $H$ and $K$ are points on
$AC$ such that $AH:HK:KC=2:3:1$. The line $DG$ meets $FH$ and $EK$ at $P$ and $Q$
respectively. Find $DP:QG$.
The suggested answer says that this problem can be solved by "mass point" techniques.

To find $DP:PG$, one may assign a mass of $20$ to $A$ and split the masses at $B$ and $C$. Using the ratios $AD:DB$ and $AH:HC$, the partial masses at $B$ and $C$ towards $A$ will be $8$ and $10$ respectively.
One can then use the ratio $BF:FG:GC$ to work out the partial masses at $B$ towards $C$ and $C$ towards $B$ to be $3.2$ and $2.8$(here I believe that it should be a typo, it should be $12.8$)respectively. Now considering $DG$, the masses at $D$ and $G$ will be $28$ and $16$ respectively,
which shows that $DP:PG=4:7$. One can then find $DQ:QG$ in a similar way.

My problem:
I am new to the "mass point" technique. I understand the first paragraph but don't understand the second paragraph. I only know the mass of $B$ towards $A$ and mass of $C$ towards $A$. How do I know that the partial masses at $B$ towards $C$ and $C$ towards $B$ are $3.2$ and $12.8$ respectively?
Reference:
Question 11 From https://www.hkage.org.hk/file/competitions/4942/Prelim2019.pdf
Official Solution: https://www.hkage.org.hk/file/competitions/5615/Prelim2019_sol.pdf
 A: Simply using vectors
$$ D = \frac{2}{7}B+\frac{5}{7}A $$
$$ E = \frac{5}{7}B+\frac{2}{7}A $$
$$ H = \frac{2}{3}A+\frac{1}{3}C $$
$$ K = \frac{1}{6}A+\frac{5}{6}C $$
$$ F = \frac{5}{6}B+\frac{1}{6}C $$
$$ G = \frac{1}{3}B+\frac{2}{3}C $$
The point $P$ is a convex combination of $H$ and $F$, $\lambda H+(1-\lambda)F$, but also a convex combination of $D$ and $G$, $\mu D+(1-\mu)G$, so
$$ P = \frac{2\lambda}{3}A+\frac{\lambda}{3}C+\frac{5(1-\lambda)}{6}B+\frac{(1-\lambda)}{6}C = \frac{2\mu}{7}B+\frac{5\mu}{7}A+\frac{1-\mu}{3}B+\frac{2(1-\mu)}{3}C$$
and $\mu_P=\frac{42}{71}$. Similarly for $Q$
$$Q=\frac{5\lambda}{7}B+\frac{2\lambda}{7}A + \frac{1-\lambda}{6}A+\frac{5(1-\lambda)}{6}C = \frac{2\mu}{7}B+\frac{5\mu}{7}A+\frac{1-\mu}{3}B+\frac{2(1-\mu)}{3}C $$
and $\mu_Q=\frac{4}{13}$. It follows that $QG=\frac{4}{17}DG$ and $DP=\frac{42}{113}DG$ and $\frac{DP}{QG}=\frac{357}{226}$.
Maybe I messed up some computations (no, I haven't, I checked this both in Mathematica and Geogebra) but the technique in itself is pretty straightforward.
