In terms of Rydberg constant R, the shortest wavelength in Balmer series of hydrogen atom spectrum will have wavelength

Explanation

Suppose the centre of the sphere 1,2 and 3 are O, A and B respectively and having radius R metre each of it.
Mass of the sphere-1 = `m_(1)=m` and coordinates of its centre O are,
`vec(r_(1))=(0,0,0)m`.
Mass of the sphere-2 = `m_(2)=m` and coordinates of its centre A are,
`vec(r_(2))=(2R,0,0)` metre
Mass of the sphere-3 = `m_(3)=m` and coordinates of its centre B are,
`vec(r_(3))=(2Rcos60^(@),2Rsin60^(@),0)`
`=(2Rxx(1)/(2),2Rxx(sqrt(3))/(2),0)`
`therefore vec(r_(3))=(R,sqrt(3)R,0)` metre
`vec(r_(cm))=(m_(1)vec(r_(1))+m_(2)vec(r_(2))+m_(3)vec(r_(3)))/(m_(1)+m_(2)+m_(3))`
`therefore vec(r)_(cm)=(m(0,0,0)+m(2R,0,0)+m(R,sqrt(3R),0))/(m+m+m)`
`=(m[((3R,sqrt(3)R,0))])/(3m)`
`=(3R,sqrt(3)R,0)/(3)`
`therefore vec(r)_(cm)=(R,(R )/(sqrt(3)),0)` metre