jmc

We use similar triangles: $△BPH\sim △APC$ since they are both right triangles and the angles at $A$ and $B$ are each complementary to $\angle C$, and thus congruent. Similarly, $△AQH\sim △BQC$. We know that $HP=5$ and $HQ=2$, so we have the ratios $$\frac{BP}{5}=\frac{AH+5}{PC}$$and $$\frac{AQ}{2}=\frac{BH+2}{QC}.$$Cross-multiplying and then subtracting the second equality from the first yields $$(BP)(PC)-(AQ)(QC)=5(AH)+25-2(BH)-4.$$But $△BPH\sim △AQH$, so $BH/5=AH/2$, so $5(AH)-2(BH)=0.$ Hence our desired answer is simply $25-4=\boxed{21}$.