Three concentric spherical metallic shells A, B and C of radii $a,b$ and $c(a<b<c)$ have surface charge densities $\sigma ,-\sigma$ and $\sigma ,$ respectively.
If $V_A,V_B$ and $V_C$ are potential of shells A, B and C, respectively, match the columns
$\begin{array}{cc}\text{Column A}& \text{Column B}\\ \text{a.}{V}_A& \text{i.}\frac{\sigma }{{\epsilon }_0}\left[\frac{a^2-b^2+c^2}{c}\right]\\ \text{b.}{V}_B& \text{ii.}\frac{\sigma }{{\epsilon }_0}[\frac{a^2}{b}-b+c]\\ \text{c.}{V}_C& \text{iii.}\frac{\sigma }{{\epsilon }_0}[a-b+c]\end{array}$

The correct option is A a − iii, b − ii, c − i
Let us consider the general case of a metallic shell in which we consider three points x, y and z which lies inside the surface, on the surface and outside the surface respectively.
So we, know in case of metallic sphere,
potential inside the sphere = potential on the surface of sphere
$⟹V_x=V_y$
So, on shell A the potential due shell B and C will be same as that on their surfaces
$⟹V_A=\frac{k\sigma 4\pi a^2}{a}-\frac{k\sigma 4\pi b^2}{b}+\frac{k\sigma 4\pi c^2}{c}$ (we know Charge = $\sigma \times 4\pi r^2$)
$⟹V_A=\frac{\sigma }{{ϵ}_0}[a-b+c]$
IIy $V_B=\frac{k\sigma 4\pi a^2}{b}-\frac{k\sigma 4\pi b^2}{b}+\frac{k\sigma 4\pi c^2}{c}$
$⟹V_B=\frac{\sigma }{{ϵ}_0}[\frac{a^2}{b}-b+c]$
$V_C=\frac{k\sigma 4\pi a^2}{c}-\frac{k\sigma 4\pi b^2}{c}+\frac{k\sigma 4\pi c^2}{c}$
$⟹V_C=\frac{\sigma }{{ϵ}_0}\left[\frac{a^2-b^2+c^2}{c}\right]$