Are you ready for the AIME?

since the 5 members of each plant committee are distinct we get that the number of arrangement of sittings is in the form N*(5!)^3 because for each M,V,E sequence we have 5! arrangements within the Ms, Vs, and Es. which results in a number above

Since after each planet, only members of another planet can follow, we simply count the lengths of the blocks adding up to ten. These blocks must be of the form MVE with a certain number of M's,V's,and E's,we consider a few different cases:
$1$ . One block of five people- There is only one way to arrange this so ${1^3}=1.$
$2$ . Five blocks of one person - There is also only one way to arrange this so we get ${1^3}=1.$
$3$ . Two blocks - There are two cases: 4+1 and 3+2. Each of these can be arranged two ways so we get ${(2+2)^3}=64.$
$4$ . Three blocks - There are also two cases: 3+1+1 and 2+2+1.Each of these can be arranged three ways giving us ${(3+3)^3}=216.$
$5$ . Four blocks - There is only one case: $2+1+1+1$ . This can be arranged four ways giving us ${4^3}=64.$
Combining all these cases, we get $1+1+64+64+216= \boxed{346}$

Not my own solution