Two equal and oppositely charged particles are kept some distance apart from each other. A spherical surface having radius equal to separation between the particles and concentric with their mid point is considered. Then:

A system of two equal but oppositely charged particles is called electric dipole. Suppose separation between the particles is equal to $2l$, then a spherical surface, concentric with centre of the dipole and having radius $2l$ is to be considered as shown in the figure. If a point $P$ be considered such that the radius passing through it, makes angle $\theta$ with the dipole axis as shown in the figure, the moment of electric dipole can be resolved into the components:$E$

(i) $p_1$, along the radius and (ii) $p_2$, normal to the radius.

Let magnitudes of electric field strength, at $P$, due to these components be $E$, and $E₂$, their directions will be as shown in the figure.

Resultant of these two can be normal to the surface only when $E₂$ is zero. It is possible only when $p₂$ is equal to zero. Hence, $\theta$ should be equal to either $0$ or $\mathrm{\pi }$. Hence, the option $\left(1\right)$is correct.

Resultant electric field can be zero only when both of the electric fields , and $E₂$ are equal to zero. It can lake place only if both the components $p_1$, and $p_2$ of dipole moment are zero. But it is impossible. Hence, resultant electric field cannot be equal to zero. Hence, the option $\left(2\right)$ is correct.

Potential will be equal to zero at those points which are equidistant from both the charged particles. All such points lie in a plane, normal to axis of the dipole and passing through its centre (means perpendicular bisector of the dipole). Section of the spherical surface by this plane forms a circle. Hence, the option $\left(3\right)$ is correct.

Net electric flux through a closed surface is directly proportional to the net charge enclosed within the face. But it is equal to zero. Hence, the net flux through the surface is also equal to zero. It means, the option  $\left(4\right)$ is also correct.