Reference point in Electric Potential I have a conceptual problem regarding the reference point we use to find electric potential.

The problem is one of the Griffiths Electrodynamics book (Problem 2.52 of 4th edition book), which asks us to find electric potential due to two long wire having charge densities $+\lambda$ and $-\lambda$ parallel to $x$-axis and asking us to find potential in $xyz$ space.

Note - I know that due to symmetry it is obvious that electric potential is to be evaluated in $y$-$z$ plane and infinity do not serve as a good reference point.
My doubt is, why for two different long wires, reference point should be same to evaluate the total electric potential due to both wires. What if i select different reference point for each wire and find the total electric potential due to both wires by superposition principle. Or is it necessary that reference point should be same.
Edit- This doubt came to me as most of the solution I found in Internet and Youtube, tends to take same reference point without explaining why they take same reference point.
 A: Because of $\vec{\nabla} \times \vec{E} = 0$, a static electric field can always be written in the form $\vec{E} = - \vec{\nabla} \phi$. The scalar potential $\phi$ is only determined up to some arbitrary constant $C$: The potential $\bar{\phi}(\vec{x}) =\phi(\vec{x}) +C$ describes the same electric field $\vec{E}(\vec{x})$ and thus the same physics. Choosing $\phi(\vec{x}_0)=0$ at some arbitrary "reference point" $\vec{x}_0$ is a pure convention. Note that the potential difference $\phi(\vec{x}_2)-\phi(\vec{x}_1)=\int_2^1 d \vec{x} \cdot E(\vec{x})$ remains unaffected by this choice.
In your specific example, you may, of course, determine $\phi_1$ (from wire 1) and $\phi_2$ (from wire 2) independently (with any conventions you like). Appealing to the superposition principle gives you a possible potential $\phi = \phi_1 +\phi_2$ for the field generated by both wires. But $\bar{\phi} = \phi_1 +\phi_2 + C$ is just as good.
Edit: In view of the comments of A.S. below, let me even be more explicit, taking the specific example of the two wires. Let us assume that the first wire is placed in the $x-$axis with charge density $\rho_1(x,y,z) = \lambda \delta(y) \delta(z)$ and the second one parallel to the $x-$axis with charge density $\rho_2(x,y,z)=-\lambda \delta(y-y_0) \delta(z-z_0)$. Suppose you have found a solution of $\vec{\nabla} \cdot \vec{E}_1 = \rho_1$ from symmetry arguments and Gauss' integral theorem, you can easily obtain the associated scalar potential
$\phi_1(x,y,z) =-\frac{\lambda}{2 \pi}\log\left(\frac{\sqrt{y^2 + z^2}}{r_1}\right)$,
where $r_1$ is an arbitrary (integration) constant (dimension of a length). The arbitrariness in the choice of $r_1$ reflects the fact that you are free to choose $\phi_1(x_1,y_1,z_1)=0$ at some arbitrary "reference point" $(x_1,y_1,z_1)= (x_1, r_1 \cos \alpha,r_1 \sin \alpha)$. (However, in view of the comments below, the notion of a "reference point" seems to be more confusing than enlightning.)
Now you repeat this procedure for the scalar potential of wire 2,
$\phi_2(x,y,z) = \frac{\lambda}{2 \pi} \log \left(\frac{\sqrt{(y-y_0)^2+(z-z_0)^2}}{r_2}\right)$,
where $r_2$ is again an arbitrary integration constant.
You have now obtained solutions of the equations $\Delta \phi_1 = -\rho_1$ and $\Delta \phi_2 = - \rho_2$. Because of the linearity of the Laplace operator $\Delta$, the sum $\phi = \phi_1 + \phi_2$ is the desired solution of $\Delta \phi = -(\rho_1 + \rho_2)$ given by
$\phi(x,y,z) = -\frac{\lambda}{2 \pi} \log \left( \frac{\sqrt{y^2+z^2}}{ \sqrt{(y-y_0)^2+(z-z_0)^2}} \right)+ C$.
If you whish, you may now use the remaining free constant $C$ to achieve $\phi(x_1,y_1,z_1)=0$ at your preferred "reference point". But this does not affect the physical field $\vec{E} = - \vec{\nabla} \phi$.
