, 2022

Solution:
Let $T$ be the foot of the $A$-altitude of $ABC$. Recall that $BT=5$ and $CT=9$.
Let $T'$ be the foot of the $P$-altitude of $PQR$. Since $T'$ is the midpoint of the possibilities for $P$, the answer is
$$\sum_{P} d(P, BC)=2 d\left(T', BC\right)$$
Since $T'$ splits $QR$ in a $5:9$ ratio, we have
$$d\left(T', BC\right)=\frac{9 d(Q, BC)+5 d(R, BC)}{14}$$
By similar triangles, $d(Q, BC)=QB=12 \cdot \frac{14}{9}$, and similar for $d(R, BC)$, giving $d\left(T', BC\right)=24$, and an answer of $48$.

$$d\left(P_{1}, BC\right)+d\left(P_{2}, BC\right)=2 d\left(T', BC\right)$$
Now, notice that since $\triangle PQR \sim \triangle ABC$, we have $QT':T'R=BT:TC$, so $TT'\parallel BQ\parallel CR$, implying that $A \in TT'$.
However, we recall a well-known fact that $A$ is the midpoint of $TT'$ (can be proven by simple similar triangles). Thus, $d\left(T', BC\right)$ is equal to two times the altitude from $A$ to $BC$. Hence, the answer is four times the altitude from $A$ to $BC$, which is $48$.

Solution:
As in the previous solution, let $T$ be the foot from $A$ to $BC$, let $T'$ be the foot from $P$ to $QR$, and recall that
$$d\left(P_{1}, BC\right)+d\left(P_{2}, BC\right)=2 d\left(T', BC\right)$$
By similar triangles, $d(Q, BC)=QB=12 \cdot \frac{14}{9}$, and similar for $d(R, BC)$, giving $d\left(T', BC\right)=24$, and an answer of $48$.