Hint: \(\oint_{{S}} {E} .{d}{s}=\dfrac{{q}}{\varepsilon_0} =\phi\)

Explanation: \(\int_S E.ds = \phi\) represents electric flux over the closed surface. 
In general, \(\oint_{{S}} {E} .{d}{s}\) represents the algebraic sum of the number of flux lines entering the surface and the number of flux lines leaving the surface. When \(\int_S E.ds = 0,\) it means that the number of flux lines entering the surface must be equal to the number of flux lines leaving it. Gauss's law also states that \(\oint_{{S}} {E} .{d}{s}=\dfrac{{q}}{\varepsilon_0}\) where \(q\) is the total charge enclosed by the gaussian surface. When \(\int_S E.ds = 0,\)it means that the net charge enclosed by the surface must be zero. Therefore, all other charges must necessarily be outside the surface. This is because charges outside the surface do not contribute to the electric flux.
Therefore, statements (c) and (d) are valid statements.
Hence, option (3) is the correct answer.