The given equation can be balanced using ion-electron method as shown below.

${\mathrm{Fe}}^{2+}\to {\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-}$         -oxidation Half reaction

$6{\mathrm{e}}^{-}+14{\mathrm{H}}^{+}+{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\to 2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$ -Reduction Half reaction

The overall reaction is 

$6{\mathrm{Fe}}^{2+}+14{\mathrm{H}}^{+}+{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\to 6{\mathrm{Fe}}^{3+}+2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$

The values $\mathrm{X}=14,\mathrm{Y}=2 \mathrm{and} \mathrm{Z} = 7$

$(\mathrm{X}+ \mathrm{Y} +\mathrm{Z}) = 14 + 2 +7$

$= 23$