Let the incircle touch $AB$ at $Z$ and $AC$ at $W.$ Tangent-chord angles at $Y$ and $X$ show $\angle AYP = \angle YXC;$ use the law of sines in triangles $APY$ and $ABX$

$AZ$ is the harmonic mean of $AP$ and $AB:$ $\frac{2}{AZ} = \frac{1}{AP} + \frac{1}{AB},$ and likewise $\frac{2}{AW} = \frac{1}{AQ} + \frac{1}{AC}$ with $AW = AZ$

Let $\omega$ touch $\overline{AB}$ at $Z$ and $\overline{AC}$ at $W,$ and set $\alpha = \angle BAX$ and $\beta = \angle AXC.$ The tangent-chord angle between $PQ$ and chord $XY$ equals the one between $BC$ and $XY,$ so $\angle QYX = \angle YXC = \beta,$ and vertical angles give $\angle AYP = \beta.$ In triangle $APY$ the law of sines gives $PY = AP\,\frac{\sin\alpha}{\sin\beta},$ and by equal tangents $PZ = PY,$ so $\frac{AZ}{AP} = 1 + \frac{PY}{AP} = 1 + \frac{\sin\alpha}{\sin\beta}.$ In triangle $ABX,$ since $\angle AXB = 180^\circ - \beta,$ similarly $BX = AB\,\frac{\sin\alpha}{\sin\beta},$ and $BZ = BX$ gives $\frac{AZ}{AB} = 1 - \frac{\sin\alpha}{\sin\beta}.$

Adding the two relations, $\frac{AZ}{AP} + \frac{AZ}{AB} = 2,$ so with $AP = 3$ and $AB = 7$ we get $AZ\left(\frac{1}{3} + \frac{1}{7}\right) = 2,$ hence $AZ = \frac{21}{5}.$ The identical argument on side $AC$ (using $\angle XAC$ in triangles $AQY$ and $ACX$) gives $\frac{AW}{AQ} + \frac{AW}{AC} = 2,$ and $AW = AZ = \frac{21}{5}$ by equal tangents from $A.$ Therefore $$\frac{1}{AQ} = \frac{10}{21} - \frac{1}{8} = \frac{59}{168},$$ so $AQ = \frac{168}{59}$ and $m + n = 168 + 59 = 227.$