JovenlyCosmo

Sorry for the late posting (understatement).

Anyways, you can imagine this question as a group of 5 people and 2 other people - that group of 5 people will always sit next to each other, such that Pierre/Rosa/Thomas will be in the first, third, and fifth position of this group of 5 people. Now, you have 2 positions left, and since there are 4 people left, you can calculate this as ${4 \choose 2}=6$. Also, their are $3!=6$ ways to have the 3 people moved. Multiplying writing these numbers, you get $36$. Therefor, you have 36 ways to arrange the group of 5 (whom you can now treat as a single person!). On the other side of the round table, you have 2 people that can be arranged in $2!=2$ ways, so you have $36*2=72$, and then multiplying by 3 gives us $72*3!=72*6=432$. However, since this is a round table, you must divide by 3 since you can rotate the table as 

A

B           C

or

B

C            A

or

C

A          B,

Giving us a total of 3 ways you overcounted the original $432$, so you must divide by 3 to get $\frac{432}{3}=\boxed{144}$.

(However, you divide by 3 because this question only asks for rotations, but other questions ask for reflections/flips, in which case divide by an additional 2.)