One electron from one 's' orbital and two electrons from 'p' orbital overlaps with 'p' orbital electrons of Flourine to form three $\mathrm{sp}^{2}$ hybridised bonds. So, $\mathrm{BF}_{3}$ has sp $^{2}$ hybridisation.

Flourine (a) But in $\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{NH}_{3}$ and $\mathrm{P} \mathrm{Cl}_{3}$, the hybridisation involves one electron from 's $^{\prime}$ orbital, 3 electrons for ${ }^{\prime} \mathrm{p}^{\prime}$ orbitals in central orbital forming $\mathrm{sp}^{3}$ hybridisation.