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

${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

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