The third ionisation enthalpy of Lutetium (Lu) is anomalously low....

The atomic number of Lutetium (Lu) is $71$.

The electronic configuration of Lu is $[\text{Xe}] 4f^{14} 5d^{1} 6s^{2}$.

During the first and second ionisations, the two $6s$ electrons are removed, forming $\text{Lu}^{2+}$ with the configuration $[\text{Xe}] 4f^{14} 5d^{1}$.

The third ionisation involves the removal of the single $5d$ electron. This results in the formation of $\text{Lu}^{3+}$ with the configuration $[\text{Xe}] 4f^{14}$.

Since the resulting $4f^{14}$ configuration is completely filled and highly stable, the removal of the third electron requires less energy. Thus, the third ionisation enthalpy of Lu is anomalously low.

Answer: the attainment of a highly stable completely filled $4f^{14}$ configuration