Solution:

Let $E$ be the extouch point on $BC$, let $I$ be the incenter, and $D'$ the reflection of $D$ over $I$. Note that $DE = AC - AB$, so $DP \cdot DQ = DE^2$. Now let $F$ be the reflection of $E$ across $D$. The length condition implies $(E, F; P, Q)$ is a harmonic bundle. We also know $XY$ is the perpendicular bisector of $DE$, so the midpoint $M$ of $D'E$ lies on $XY$. But then $IM \parallel BC$, so $IM \perp XY$, and $M$ is the midpoint of $XY$. Since $A, D', E$ are collinear, this means $AE$ bisects $XY$.

Now consider projecting $(E, F; P, Q)$ onto $XY$. $P$ and $Q$ are taken to $X$ and $Y$, while $E$ is taken to the midpoint of $XY$. Thus, $F$ is taken to the point at infinity, so $AF \perp BC$. Now since $D$ is the midpoint of $EF$, we see that $AF = 2 DD'$, or $h_a = 2r$, where $h_a$ is the height from $A$ and $r$ is the inradius. But $\frac{1}{2} a h_a = \frac{a + b + c}{2} r$, so $a = \frac{a + b + c}{2}$, or $3a = b + c$, as desired.