Three particles carrying charge of equal magnitude are fixed at the vertices of an equilateral triangle ABC. The electric field at the centroid G of the triangle is directed along $\overrightarrow{\text{GB}}$. Which of these options is/are correct?

It is clear that option (A) is correct

Let the particles carry charge $Q$ each and let the side length of the triangle be $a$ Then, initially, the field at G is $E_{G1}=2\left(\frac{Q}{4\pi {\epsilon }_0{\left(\frac{a}{\sqrt{3}}\right)}^2}\right)=\frac{3Q}{2\pi {\epsilon }_0{a}^2}$

After the particle at C is moved to a very large distance from the other particles, the electric field at G becomes $E_{G2}=\sqrt{3}\left(\frac{Q}{4\pi {\epsilon }_0{\left(\frac{a}{\sqrt{3}}\right)}^2}\right)=\frac{3\sqrt{3}Q}{4\pi {\epsilon }_0{a}^2}$ So, $E_{G2}=\frac{\sqrt{3}}{2}{E}_{G1}$

If (starting from the initial situation) the particle at B is moved to G, the electric field at B due to the particles at A and C is $E_{B1}=\sqrt{3}\left(\frac{Q}{4\pi {\epsilon }_0{a}^2}\right)$ (directed along $\overline{GB}$ ),  and the electric field at B due to the particle at G is $E_{B2}=\frac{Q}{4\pi {\epsilon }_0{\left(\frac{a}{\sqrt{3}}\right)}^2}=\frac{3Q}{4\pi {\epsilon }_0{a}^2}$ (directed along $\overline{GB}$)

Since $E_{B2}>E_{B1},$the net field at B is directed along $\overline{GB}$

We can dedude easily that in the initial situation, the electric field at A is directed parallel to $\overline{CB}$