jmc

After Pierre sits, we can place Rosa either two seats from Pierre (that is, with one seat between them) or three seats from Pierre. We tackle these two cases separately:

Case 1: Rosa is two seats from Pierre. There are $2$ such seats. For either of these, there are then four empty seats in a row, and one empty seat between Rosa and Pierre. Thomas can sit in either of the middle two of the four empty seats in a row. So, there are $2\cdot 2=4$ ways to seat Rosa and Thomas in this case. There are then $4$ seats left, which the others can take in $4!=24$ ways. So, there are $4\cdot 24=96$ seatings in this case.

Case 2: Rosa is three seats from Pierre (that is, there are $2$ seats between them). There are $2$ such seats. Thomas can't sit in either of the $2$ seats directly between them, but after Rosa sits, there are $3$ empty seats in a row still, and Thomas can only sit in the middle seat of these three. Once again, there are $4$ empty seats remaining, and the $4$ remaining people can sit in them in $4!=24$ ways. So, we have $2\cdot 24=48$ seatings in this case.

Putting our two cases together gives a total of $96+48=\boxed{144}$ seatings.