For $[Fe(NH_3)_6]^3+$, you can answer as following:
- Oxidation state of Fe in this complex is '+3', making it a $3d^5$ configuration, with 5 unpaired electrons occupying each degenerate orbital.
- Since $NH_3$ is a strong field ligand, electron pairing occurs disobeying Hund's rule. Now, the configuration becomes $3d^5$ with two orbitals containing 4 paired electrons, and 1 one unpaired in one degenerate orbitals, leaving two vacant 3d orbitals.
- Now, hybridisation occurs with two 3d, one 4s, and three 4p vacant atomic orbitals forming six $d^2sp^3$ hybrid orbitals arranged in an octahedral geometry.
- The lone pair on each $NH_3$ molecule will now form coordinate covalent bond with six hybrid orbitals.
While ligand strength can be applied in VBT, it might bring some incorrect results sometimes regarding magnetic moment.
This answer might seem correct given it agrees with CFT but this doesn't with few other complexes.
Take, for example, Hexacyannonickelate (II) $[Ni(CN)_6]^4-$ for example.
VBT argues that it is a diamagnetic species because all the 8 electrons undergo pairing.
While CFT argues that the configuration will be $t_g^6 e_g^2$ with 2 unpaired electrons in $e_g$ orbitals, meaning it is paramagnetic in nature.
VBT fails to explain magnetic properties of few complexes and geometry using hybridisation, and hence CFT is preferred over VBT.