There are 3 cases here: -
You can select: -
1. One French, One Spanish and One German in 4C1(French) X 6C2 (S + G) = 4 X 15 = 60
2. Two French and one Spanish or one German in 6C1 ( S / G) X 4C2 (F)
= 36
3. All the 3 french in 4C3 ways = 4
60+36+4 = 100
Alternately: -
You can select any 3 from the 10 in 10C3 ways ---------(1)
You can select persons other than French in 6C3 ways-------(2)
So the number of ways in which at least a french will be included in the committee is 10C3-6C3 = 120 - 20 = 100 ways