Solution:

Denote the colors by $1, 2, 3$. Notice that all three colors must be used. This is true because if only two colors were used, the vertex coloring would be of the type $121212 \ldots$ which is impossible, since $17$ is odd (this would make two adjacent vertices the same color). Hence there are three consecutive vertices colored (without loss of generality) $1, 2, 3$, respectively. (For otherwise, if $3$ consecutive vertices had coloration of the form $a b a$, the next three vertices [overlapping two vertices] would have to be colored $b a b$, etc., forcing the pattern $a b a b a b \ldots$ which is not possible with an odd number of vertices.)

Thus we have $4$ cases depending on the colors of the vertices adjacent to this segment: $21231$, $21232$, $31231$, $31232$. It is easy to see that the desired construction can be done in each case; the figure below illustrates this.