Official solution
F1 \[ \to \] 3 members
F2 \[ \to \] 3 members
F3 \[ \to \] 4 members
Total arrangements of three families = 3!
Arrangement between members of F1 family = 3!
Arrangement between members of F2 family = 3!
Arrangement between members of F3 family = 4!
\[ \therefore \] Total numbers of ways can they be seated so that the same family members are not separated
= 3! \[ \times \] 3! \[ \times \] 3! \[ \times \] 4!
= (3!)3.(4!)