## Calculus 8th Edition

$r=(\sqrt 3-\dfrac{3}{2})m$
Suppose $1$ ball has center at $(r,r,r)$ $(x-r)^2+(y-r)^2+(z-r)^2=r^2$ $\lt \frac{m}{2},\frac{m}{2},\frac{m}{2} \gt =(1+\frac{2}{\sqrt 3}) \cdot \lt r,r,r \gt$ $\frac{m}{2}=r(\frac{\sqrt 3+2}{\sqrt3})$ $r=\dfrac{\sqrt 3}{4+2\sqrt3}m$ Hence, $r=(\sqrt 3-\dfrac{3}{2})m$