Find the lowest possible empirical formula in an arrangement of unit cell where $\mathrm{A}$ atoms are present at comers and alternate face centers, $\mathrm{B}$ atoms are present at alternate edge centers and $\mathrm{C}$ atoms are present at half of the mid of line joining opposite face centers.

Assume : Any atom present in the inner locations of the unit cell should be considered completely within the unit  cell structure.

As it is shared by eight unit cells, an atom contributes 1/8 of it at the corner. Considering that a unit cell has eight edges. Therefore, the total combination of$\mathrm{A}$ present in the corners is

$\Rightarrow \frac{1}{8}\times 8=1$.

alternative face center's also have it. $\mathrm{A}$ will therefore be found on any two faces that are opposite one another. The atom's contribution is 1/2 at faces.
As a result, the total contribution of A present at alternate faces is $\Rightarrow \frac{1}{2}\times 2=1$. 

The sum of the A atoms is therefore $1+1=2.$

At the alternate edge Centre, there is a $\mathrm{B}$ atom. $\mathrm{B}$ may be found on 4 of the edge center's. Consequently, $\mathrm{B}\text{'}\mathrm{s}$overall contribution is $\mathrm{B}$ $\Rightarrow \frac{1}{4}\times 4=1$

On the line connecting the two face center's, where the $\mathrm{C}$ atom is located, there are actually two atoms. There are 3 lines joining the opposing face because there are 6 faces. There are 2 atoms in a line. Consequently, there are six atoms in all. The cell contribution is 1, just like inside the unit. Consequently, the $\mathrm{C}$ atom's overall contribution is $1\times 6=6$

The empirical formula is therefore ${\mathrm{A}}_2{\mathrm{BC}}_6.$