 Solution

For the regular octahedron, repulsion from the six oxygen ligands raises the two $e_g$ orbitals, $d_{z^2}$ and $d_{x^2-y^2}$, above the three $t_{2g}$ orbitals, $d_{xy},d_{xz},d_{yz}$. The latter point between the metal-ligand axes and experience less direct repulsion. The crystal field splitting is therefore a lower triplet and an upper doublet.

For the $3d^2$ V${}^{3+}$ ion, Hund's rules put the two electrons into different $t_{2g}$ orbitals with parallel spins, giving $S=1$. There are three choices for the unoccupied $t_{2g}$ orbital, so the orbital ground state is triply degenerate. Its spin degeneracy is separate. This orbital degeneracy makes a symmetry-lowering Jahn-Teller effect possible.