Two equal spheres of mass $\mathrm{m}$ are in contact on a smooth horizontal table. A third identical sphere impinges symmetrically on them and reduces to rest. then: 

From momentum conservation

$\mathrm{mu}=\mathrm{mv} \cos 30 + \mathrm{mv} \cos 30 ; \mathrm{v}=\frac{\mathrm{u}}{\sqrt{3}}$

So, choice (3) is correct. For an oblique collision, we have to take components along normal i.e., along $\mathrm{AB}$  for sphere $\mathrm{A} \mathrm{and} \mathrm{B}$

${\mathrm{v}}_2-{\mathrm{v}}_1=\mathrm{e}\left({\mathrm{u}}_1-{\mathrm{u}}_2\right) ; \mathrm{v}-0=\mathrm{e}\left[\mathrm{u} \cos 30-0\right]$

$\mathrm{v}=\mathrm{eu} . \frac{\sqrt{3}}{2}$   ;   $\mathrm{v}=\mathrm{e} . \mathrm{v}\sqrt{3}.\frac{\sqrt{3}}{2} ; \mathrm{e}=\frac{2}{3}$

So, choice (1) is correct. Also, loss of kinetic energy

$$\begin{gathered} ∆\mathrm{K}=\frac{1}{2}{\mathrm{mu}}^2-2\left(\frac{1}{2}{\mathrm{mv}}^2\right) \\ =\frac{1}{2}{\mathrm{mu}}^2-2\frac{1}{2}\mathrm{m}{\left(\frac{\mathrm{u}}{\sqrt{3}}\right)}^2=\frac{1}{6}{\mathrm{mu}}^2 \end{gathered}$$