\(D\) and \(E\) are the midpoints of \(B C\) and \(AC\).

\(\therefore A E=E C=a\) and \(B D=D C=a\)

\(\operatorname{In} \Delta A D C,(A D)^{2}=(A C)^{2}-(D C)^{2}\)

\(=(2 a)^{2}-(a)^{2}=4 a^{2}-a^{2}=3 a^{2}\)

\(A D=a \sqrt{3}\)

Similarly, potential at point \(D\) due to the given charge configuration is

\(V_{D}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{B D}+\frac{q}{D C}+\frac{q}{A D}\right]\)

\(=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{a}+\frac{1}{a}+\frac{1}{\sqrt{3} a}\right]=\frac{q}{4 \pi \varepsilon_{0} a}\left[2+\frac{1}{\sqrt{3}}\right].........(i)\)

Potential at point \(E\) due to the given charge configuration is \(V_{E}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{A E}+\frac{q}{E C}+\frac{q}{B E}\right]\)

\(=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{a}+\frac{1}{a}+\frac{1}{a \sqrt{3}}\right]=\frac{q}{4 \pi \varepsilon_{0} a}\left[2+\frac{1}{\sqrt{3}}\right].........(ii)\)

From the \((i)\) and \((ii)\), it is clear that

\(V_{D}=V_{E}\)

The work done in taking a charge \(Q\) from \(D\) to \(E\) is

\(W = Q({V_E} - {V_D}) = O\) \((\because \,{V_D} = {V_E})\)