Calculate length of one side given all other sides in a triangle. 
All the lengths in red had been given in the problem statement. The task is to find $BC$. The green ones were calculated from the information in red.
Here's two ways I tried to solve it (and I would like feedback):



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The green line is drawn parallel to $QC$. Suppose, the other unlabelled end of the green line be point $R$. So the line can be called $PR$.   
In $\triangle BPR$,
$PC=QC=9, \space\space PC=3\times AP\\PB=6, \space\space PB=3 \times AQ\\\therefore \triangle BPR \sim \triangle APQ\\BD=4\times3, \space\space \boxed{BD=12}$
In quadrilateral $PQCD$, opposite sides are parallel to each other and it is therefore a parallelogram. So $DC=4$.
$BC=BD+DC\\BC=12+4\\\boxed{BC=16}$




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In $\triangle ABC$, $AB=2+6=2(1+3)=\boxed{2\times 4}\\AC=3+9=3(1+3)=\boxed{3\times 4}\\BC=?$
In $\triangle APQ$, $AP=\boxed{2}\\AC=\boxed{3}\\PQ=4$
$\because \triangle ABC\sim \triangle APQ,\\BC=PQ\times4=4\times4=16,\\\boxed{BC=16}$


Your question doesn't mention $PQ$ is parallel to $BC$ but your answer does. 
So assuming that line $PQ$ is parallel to $BC$ then by similar triangles 
( Considering the triangles $APQ$ and $ABC$)
$$\frac {AP}{AB}= \frac {AQ}{AC}=\frac {PQ}{BC}$$
Hence we get $$\frac {3}{12}=\frac {4}{BC}$$
$$\Rightarrow BC=16$$
