1. Potential due to one infinitely long charged wire

For an infinite line charge of linear charge density $\lambda$, the electric potential at perpendicular distance $r$ is

$V = \frac{\lambda}{2\pi \varepsilon_0}\ln\!\left(\frac{R}{r}\right)$

where $R$ is a reference distance. Thus, apart from an additive constant,

$V \propto -\ln r$

2. Perpendicular distances from the three axes

The three wires lie along:

• $x$-axis
• $y$-axis
• $z$-axis

So, for a point $(x,y,z)$, the perpendicular distances are:

• from the $x$-axis: $r_x = \sqrt{y^2+z^2}$
• from the $y$-axis: $r_y = \sqrt{x^2+z^2}$
• from the $z$-axis: $r_z = \sqrt{x^2+y^2}$

3. Total potential at $(x,y,z)$

Since potential is a scalar, total potential is the sum:

$V = \frac{\lambda}{2\pi\varepsilon_0}\left[\ln\left(\frac{R}{r_x}\right)+\ln\left(\frac{R}{r_y}\right)+\ln\left(\frac{R}{r_z}\right)\right]$

Combining logarithms,

$V = \frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{R^3}{r_x r_y r_z}\right)$

For an equipotential surface, $V =$ constant. Therefore,

$r_x r_y r_z = \text{constant}$

Substitute the distances:

$\sqrt{y^2+z^2}\,\sqrt{x^2+z^2}\,\sqrt{x^2+y^2} = \text{constant}$

Squaring both sides,

$(y^2+z^2)(x^2+z^2)(x^2+y^2)=\text{constant}$

This matches:

$\left(x^2+y^2\right)\left(y^2+z^2\right)\left(z^2+x^2\right)=\text{constant}$

4. Check options

• A: $\left(x^2+y^2\right)\left(y^2+z^2\right)\left(z^2+x^2\right)=\text{constant}$ ✔️
• B, C, D: These do not arise from the product of perpendicular distances from the three axes. ✖️

Hence the correct option is A.

5. Comparison with stored answer

Stored correct answer: A

My derived answer: A

So, they agree.