Maximum number of coplanar atoms in difluorotrimethylphosphorane In $\ce{PF2(CH3)3},$ what is the maximum number of atoms lying in one plane?

$\ce{P}$ is the central atom with the hybridization $\mathrm{sp^3d}.$ Two $\ce{F}$ atoms will form axial bonds with it, while three $\ce{CH3}$-groups form equatorial bonds. Each of these $\ce{CH3}$-groups has tetrahedral geometry, which means that one $\ce{H}$ atom lies in the same plane as the $\ce{C}$ central atom.
Thus, two $\ce{F}$ atoms, one $\ce{P}$ atom, one $\ce{C}$ atom and one $\ce{H}$ atom with it, totalling five atoms, lie in the same plane.
Is my attempt correct?
Seven is correct (the OP found this in the comments).  The methyl carbons are in the equatorial plane and each one can rotate a hydrogen atom into this plane.  In fact the hydrogen atoms fit well with this conformation so it's actually favorable.