d
\(V_{A}=\frac{1}{4 \pi \varepsilon_{0}}\left\{\frac{q_{A}}{a}+\frac{q_{B}}{b}+\frac{q_{C}}{c}\right\}\)

\(=\frac{4 \pi}{4 \pi \varepsilon_{0}}\left\{\frac{a^{2} \sigma}{a}-\frac{b^{2} \sigma}{b}+\frac{c^{2} \sigma}{c}\right\}\)

\({V_{A}=\frac{1}{\varepsilon_{0}}\left\{\frac{a^{2} \sigma}{a}-\frac{b^{2} \sigma}{b}+\frac{c^{2} \sigma}{c}\right\}} \)

\({V_{B}=\frac{1}{\varepsilon_{0}}\left\{\frac{a^{2} \sigma}{b}-\frac{b^{2} \sigma}{b}+\frac{c^{2} \sigma}{c}\right\}}\)

\(V_{C}=\frac{1}{\varepsilon_{0}}\left\{\frac{a^{2} \sigma}{c}-\frac{b^{2} \sigma}{c}+\frac{c^{2} \sigma}{c}\right\}\)

Given \(c=a+b\) If \(a=a, b=2 a\) and \(c=3 a\) for example, as \(c>b>a\)

\({V_{A}=\frac{1}{\varepsilon_{0}}\left\{\frac{a^{2} \sigma}{a}-\frac{4 a^{2} \sigma}{2 a}+\frac{c^{2} \sigma}{c}\right\}} \)

\({V_{B}=\frac{1}{\varepsilon_{0}}\left\{\frac{a^{2} \sigma}{2 a}-\frac{4 a^{2} \sigma}{2 a}+\frac{c^{2} \sigma}{c}\right\}}\)

\(V_{C}=\frac{1}{\varepsilon_{0}}\left\{\frac{a^{2} \sigma}{3 a}-\frac{4 a^{2} \sigma}{3 a}+\frac{c^{2} \sigma}{c}\right\}\)

It can seen by taking out common factors that

\(V_{A}=V_{C}>V_{B} \quad \text { i.e., } \quad V_{A}=V_{C} \neq V_{B}\)