For simplicity, let Petya be $A$, Vasya be $B$, Tolya be $C$. We can see the boundary cells as a cycle $P_1 P_2 \dots P_{6n} $ ($n=66$), where the cell $P_i$ is adjacent to the cells $P_{i-1},P_{i+1}$ (we always take indices modulo $6n$). Then, the cells $P_i, P_{i+3n}$ are symmetrical. The strategy for $B, C$ to win is simply play a "120-degree symmetric role" with $A$. More precisely, if $A$ paints a cell $P_i$, then $B$ paints a cell $P_{i+2n}$ and $C$ paints a cell $P_{i-2n}$. Observe that painting this way, after $B,C$ make their move, we have that, for every $i=1, \dots , 6n$, the cells $\{ P_i, P_{i+2n}, P_{i-2n} \} $ are all painted, or all not painted. Besides, if $P_i$ is not painted and it's forbidden to be painted, so do $P_{i+2n}, P_{i-2n}$, because if $P_i$ is symmetrical (or adjacent) to some painted cell, the by the "120-degree symmetric role", $P_{i+2n}, P_{i-2n}$ are symmetrical (or adjacent) to some painted cell. Therefore, if $A$ can play, $B,C$ can also play and they never lose. So, eventually $A$ loses.