The sum of number of lone pairs of electrons present on the central...

Number of lone pairs on the central atom=Number of hybrid orbitals of the central atom$-$Number of sigma bonds of the central atom.

Number of hybrid orbitals or

Hybridisation for a molecule is given by $= \dfrac{1}{2}\left(V + H - C + A\right)$

Where V= Number of valance electrons in central atom.

H = Number of surrounding monovalent atoms.

C= Cationic charge

A = Anionic charge.

${XeF}_{6}:$

Number of hybrid orbitals are 7.

Number of sigma bonds formed by the central atom are 6

Number of lone pairs on the central atom=7-6=1 lone pair

${XeOF}_{4}: \dfrac{8 + 4}{2} = 6$

Number of hybrid orbitals are 6

Number of sigma bonds formed by the central atom are 5

Number of lone pairs on the central atom=6$-$5=1

${XeO}_{3}$:

Number of hybrid orbitals=4

Number of sigma bonds formed by the central atom are 3

Number of lone pairs on the central atom=4$-$3=1

Hence, the total number of lone pairs on the central atom of ${XeO}_{3}, {XeOF}_{4}and{XeF}_{6}$ are 3.