Lattice vectors for rhombohedral unit cell of BiFeO3

[Iyyapparajan]

Hi,
I am doing structural relaxation for rhombohedral lattice of BiFeO3. The experimental lattice constant that we are using 5.63 and the wyckoff positions are Bi(2a) is 0, Fe(2a) is 0.221 and O(6b) is 0.538, 0.933, 0.395. I have set the lattice vectors of rhombohedral unit cell as (0,0,0), (1/3, 2/3, 2/3), (2/3, 1/3, 1/3). When i run vasp by using the following POSCAR, it shows the message in the OUTCAR file that the distance between the atoms are small.
POSCAR file is attached below :
BiFeO3
5.63
0.000 0.000 0.000
0.333 0.667 0.667
0.667 0.333 0.333
2 2 6
Direct
0.000 0.000 0.000
0.500 0.500 0.500
0.721 0.721 0.721
0.221 0.221 0.221
0.895 0.433 0.038
0.433 0.038 0.895
0.038 0.895 0.433
0.538 0.933 0.395
0.395 0.538 0.933
0.933 0.395 0.538

Are the lattice vectors given above correct ?
My question is how to calculate the lattice vectors for rhombohedral unit cell for the POSCAR file with the lattice constant of 5.63 and the wyckoff positions are Bi(2a) is 0, Fe(2a) is 0.221 and O(6b) is 0.538, 0.933, 0.395. Kindly help me.

Thanks,
Iyyappa Rajan P

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