We use here a coordinate system with the constant covariant metric:

1 | −½ | −½ |

−½ | 1 | −½ |

−½ | −½ | 1 |

In this system the following are unit vectors ±(1, 0, 0), ±(0, 1, 0), ±(0, 0, 1), ±(0, 1, 1), ±(1, 0, 1), ±(1, 1, 0).
Just now I see that the length of (1, 1, 1) is 0 (recall that Length^{2} = g_{ij}x^{i}x^{j}) and that distresses me.
Indeed this shows that the eigenvalues of g are 0, 3/2 and 3/2.

Adopt plain coordinates, x, y, z. The basis vectors used above in plain coordinates are:

(a, b, b), (b, a, b), (b, b, a) where a

a

2(√(1 − 2b

2(√(1 − 2b

√(1 − 2b

1 − 2b

b

b

16b

36b

b

b = −√(1/6) = −0.408248290463863

a = √(⅔) = 0.816496580927726

The point which is at (1, 1, 1) in oblique coordinates is at (c, c, c) in plain coordinates where c = a + 2b = 0. Oops. The puzzle pieces are not shaped as I thought!

This is obvious in retrospect. I required that the basis vectors for our oblique system be 120° from each other. In 3D three such vectors must lie in the same plane and thus cannot span the space.