Distance between nearest octahedral and tetrahedral void in FCC unit cell of edge length $8\sqrt{3}\stackrel{0}{\mathrm{A}} \mathrm{is} \mathrm{x} \stackrel{0}{\mathrm{A}},$ then x is______

Let we assume one corner of unit cell at origin so the corners of units cell will be at 
$\mathrm{A}(0,0,0),\mathrm{B}(8\sqrt{3},0,0),\mathrm{C}(8\sqrt{3},0,8\sqrt{3}),\mathrm{D}(0,0,8\sqrt{3}),\mathrm{w}(0,8\sqrt{3},8\sqrt{3}),$ $\mathrm{x}(8\sqrt{3},8\sqrt{3},8\sqrt{3}),\mathrm{y}(8\sqrt{3},8\sqrt{3},0),\mathrm{z}(0,8\sqrt{3},0)$. 

Now two nearest octahedral void which will be either on edge centre of body centre will be nearest to tetrahedral void on nearest body diagonal.
So, tenth of body diagonal $=\left(8\sqrt{3}\right)\sqrt{3}=24\mathrm{A}$ 
Let octahedral void is on AD and nearest tetrahedral void is on AX from ¼ of total AX from A
So, coordinate of octahedral centre=$(0,0,4\sqrt{3})$
Coordinate of tetrahedral centre $=(2\sqrt{3},2\sqrt{3},2\sqrt{3})$
Distance b/w these voids $=\sqrt{(2\sqrt{3}{)}^2+(2\sqrt{3}{)}^2+(2\sqrt{3}{)}^2}=6\stackrel{0}{\mathrm{A}}$