ways of selecting consecutive persons sitting at a table

You forgot the possibility that the second person can be chosen to sit two seats away from the first, and then the third person is chosen to be the one between the first and the second. This gives an additional $\frac29\cdot \frac18 = \frac1{36}$, bringing the total up to $\frac1{18}+\frac1{36} = \frac1{12}$.

For your mistake see the answer of Arthur.

A bit more concise solution:

If the first person has been chosen then yet $2$ out of $9$ must be chosen.

In $3$ of these cases the three chosen persons will sit consecutively so the probability on that is:$$\frac3{\binom92}=\frac1{12}$$