Just calculate the number of pairs(a,b) where a is a route,b is a stop and a includes b.

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Let $s$ be the number of stops and $L$ the number of lines. Now each pair of stops is covered by exactly one line, and each line has exactly three pairs of stops, so we get $3L=\frac{s\left(s-1\right)}{2}$ or $L=\frac{s\left(s-1\right)}{6}$ On the other hand, suppose the first line has stops $s_1,\ s_2,\ s_3$. There are $s-3$ remaining pairs of stops including $s_1$ and likewise for $s_2$ and $s_3$. Each line must have exactly one of $s_1$, $s_2$ or $s_3$ and each line will contain exactly two pairs of stops including one of $s_1$, $s_2$ or $s_3$. So there will be $1$ line will all $\left\{s_1,\ s_2,\ s_3\right\}$, then $\frac{s-3}{2}$ lines containing $s_1$, and likewise for $s_2$ and $s_3$, and this will cover all the lines. So we also have the relation $L=3\cdot\frac{s-3}{2}+1$. So we have $3\cdot\frac{s-3}{2}+1=\frac{s\left(s-1\right)}{6}$ or $\left(s-3\right)\left(s-7\right)=0$ If $s=3$ we have the trivial case where we have one line passing through all three stops. If $s=7$, call the seven stops $1, 2, 3, 4, 5, 6, 7$. Then from the above formula we get $L=7$ and could have for example the following arrangement: $\left\{1,\ 2,\ 3\right\}$ $\left\{1,\ 4,\ 5\right\}$ $\left\{1,\ 6,\ 7\right\}$ $\left\{2,\ 4,\ 6\right\}$ $\left\{2,\ 5,\ 7\right\}$ $\left\{3,\ 4,\ 7\right\}$ $\left\{3,\ 5,\ 6\right\}$