The number of lone pairs present on the central atom of ${\mathrm{XeF}}_2,{\mathrm{XeF}}_4$ and ${\mathrm{XeF}}_6$ are respectively $\mathrm{x},\mathrm{y}$ and $\mathrm{z}$. The product of these three coefficients is

For ${\mathrm{XeF}}_2$,

Total number of electron pair $=\frac{8+2}{2}=5$

Total number of bond pair $=2$

Total number of lone pair $=(5-2)=3=\mathrm{x}$

For ${\mathrm{XeF}}_4$,

Total number of electron pair $=\frac{8+4}{2}=6$

Total number of bond pair $=4$

Total number of lone pair $=(6-4)=2=y$

For ${\mathrm{XeF}}_6$,

Total number of electron pair $=\frac{8+6}{2}=7$

Total number of bond pair $=6$

Total number of lone pair $=(7-6)=1=z$

Product of $\mathrm{x},\mathrm{y}$ and $\mathrm{z}=3\times 2\times 1$

$=6$

Hence, The product of these three coefficients is 6.