11. 29
11. We have $56=A_{1} A_{11}=A_{1} A_{2}+A_{2} A_{5}+A_{5} A_{8}+A_{8} A_{11} \geq A_{1} A_{2}+17+17+17$, so that $A_{1} A_{2} \leq 5$. On the other hand we have $A_{1} A_{2}=A_{1} A_{4}-A_{2} A_{4} \geq 17-12=5$, so we must have $A_{1} A_{2}=5$. Similarly $A_{10} A_{11}=5$.
Next we have $A_{2} A_{7}=A_{1} A_{4}+A_{4} A_{7}-A_{1} A_{2} \geq 17+17-5=29$ on one hand, and on the other hand we have $A_{2} A_{7}=A_{1} A_{11}-A_{1} A_{2}-A_{1} A_{10}-A_{10} A_{11} \leq 56-5-17-5=29$. It follows that $A_{2} A_{7}=29$.
Remark. The scenario in the question is indeed possible. One example is $A_{i} A_{i+1}=5,7,5,5,7$, $5,5,7,5,5$ for $i=1,2, \ldots, 10$.
11. Let $A_{1}, A_{2}, \ldots, A_{11}$ be 11 points on a straight line in order, where $A_{1} A_{11}=56$. Given that $A_{i} A_{i+2} \leq 12$ for $i=1,2, \ldots, 9$ and $A_{j} A_{j+3} \geq 17$ for $j=1,2, \ldots, 8$, find $A_{2} A_{7}$.
Let $A_{1}, A_{2}, \ldots, A_{11}$ be 11 points on a straight line in order, where $A_{1} A_{11}=56$. Given that $A_{i} A_{i+2} \leq 12$ for $i=1,2, \ldots, 9$ and $A_{j} A_{j+3} \geq 17$ for $j=1,2, \ldots, 8$, find $A_{2} A_{7}$.