Electric field due to infinitely long wire

$$\overrightarrow E = - {{2k\lambda } \over r}\widehat r$$

We know, $$V = - \int {\overrightarrow E \,.\,\overrightarrow {dr} } $$

$$ = - \int {{{ - 2k\lambda } \over r}\widehat r\,.\,\overrightarrow {dr} } $$

$$v = \int {{{2k\lambda } \over r}dr = 2k\lambda \,\ln r + c} $$

Net potential due to all wires,

$$V = {v_1} + {v_2} + {v_3}$$

$$ = 2k\lambda \ln \left( {\sqrt {{x^2} + {y^2}} } \right) + 2k\lambda \ln \left( {\sqrt {{y^2} + {z^2}} } \right) + 2k\lambda \ln \left( {\sqrt {{z^2} + {x^2}} } \right) + c$$

for equipotential surface, $$v = c$$

$$ \Rightarrow 2k\lambda \left[ {\ln \left( {\sqrt {{x^2} + {y^2}} \,.\,\sqrt {{y^2} + {z^2}} \,.\,\sqrt {{z^2} + {x^2}} } \right)} \right] = c$$

$$ \Rightarrow \sqrt {({x^2} + {y^2})\,.\,({y^2} + {z^2})\,.\,({z^2} + {x^2})} = c$$

$$ \Rightarrow ({x^2} + {y^2})({y^2} + {z^2})({z^2} + {x^2}) = c$$

where, c = constant.