${V_A} = {1 \over {4\pi {\varepsilon _0}}}\left\{ {{{qA} \over a} + {{qB} \over b} + {{qC} \over c}} \right\}$

$= {{4\pi } \over {4\pi {\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over a} - {{{b^2}\sigma } \over b} + {{{c^2}\sigma } \over c}} \right\}$

${V_A} = {1 \over {{\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over a} - {{{b^2}\sigma } \over b} + {{{c^2}\sigma } \over c}} \right\}$

${V_B} = {1 \over {{\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over a} - {{{b^2}\sigma } \over b} + {{{c^2}\sigma } \over c}} \right\}$

${V_C} = {1 \over {{\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over a} - {{{b^2}\sigma } \over b} + {{{c^2}\sigma } \over c}} \right\}$

Given c = a + b.
If a = a, b = 2a and c = 3a for example, as c > b > a,

${V_A} = {1 \over {{\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over a} - {{4{b^2}\sigma } \over {2a}} + {{{c^2}\sigma } \over c}} \right\}$

${V_B} = {1 \over {{\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over {2a}} - {{4{a^2}\sigma } \over {2a}} + {{{c^2}\sigma } \over c}} \right\}$

${V_C} = {1 \over {{\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over {3a}} - {{4{a^2}\sigma } \over {3a}} + {{{c^2}\sigma } \over c}} \right\}$

It can seen by taking out common factors that
V_A = V_C > V_B i.e., V_A = V_C $\ne$ V_B