13. 1. 3 * The vertices of $\triangle A B C$ are $A(0,0), B(0,420), C(560,0)$, and a die has its six faces marked with $A, A, B, B, C, C$. A point $P_{1}=(k, m)$ is taken inside $\triangle A B C$, and points $P_{2}, P_{3}, P_{4}, \cdots$ are generated according to the following rule: If $P_{n}$ is already determined, roll the die to get a mark $L, L \in\{A, B, C\}$, then $P_{n+1}$ is the midpoint of $P_{n} L$. Given $P_{7}=(14,92)$, find: the value of $k+m$.
From the given conditions, we can determine that $P_{7}(14,92)$ can only be the midpoint of $P_{6} A$, so $P_{6}(28,184)$. Similarly, $P_{6}$ can only be the midpoint of $P_{5} A$, so $P_{5}(56,368)$. $P_{5}$ can only be the midpoint of $P_{4} B$, so $P_{4}(112,316)$. $P_{4}$ can only be the midpoint of $P_{3} B$, so $P_{3}(224,212)$. $P_{3}$ can only be the midpoint of $P_{2} B$, so $P_{2}(448,4)$. $P_{2}$ can only be the midpoint of $P_{1} C$, so $P_{1}(336,8)$. Therefore, $k+m=336+$ $8=344 . \quad$