Energies of d orbitals relative to the barycentre
The passage you quoted is perfectly correct and contains one word which you may have overlooked:


  the weighted mean of these two sets of perturbed orbitals is taken as zero [...]


In this case the weight corresponds to the degeneracy of the set. The lower-energy $\mathrm{t_{2g}}$ set contains three orbitals and the higher-energy $\mathrm{e_g}$ set contains two. If we place the $\mathrm{t_{2g}}$ set at $x \Delta_\mathrm{O}$ and the $\mathrm{e_g}$ set at $y\Delta_\mathrm{O}$, then

$$\begin{align}
y - x &amp;= 1 &amp; &amp;\text{(difference must be equal to }\Delta_\mathrm{O}) \\
\frac{3x + 2y}{3+2} &amp;= 0 &amp; &amp;(\textit{weighted}\text{ mean is equal to zero)}
\end{align}$$

and $(x,y) = (-0.4, 0.6)$.