A charge $$Q$$ is distributed over three concentric spherical shells of radii $$a$$, $$b$$, $$c$$ ($$a < b < c$$) such that their surface charge densities are equal to one another. The total potential at a point at distance $$r$$ from their common centre, where $$r < a$$, would be:

Let the common surface charge density on every shell be $$\sigma$$ (in $$\text{C m}^{-2}$$). For a spherical conducting shell of radius $$R$$, the charge $$q$$ on it is related to $$\sigma$$ by the elementary relation

$$q \;=\; \sigma \,\times\, 4\pi R^{2}.$$

Therefore, if $$q_{a},\,q_{b},\,q_{c}$$ are the charges on the shells of radii $$a,\,b,\,c$$ respectively, we have

$$\begin{aligned} q_{a} &= \sigma \, 4\pi a^{2},\\ q_{b} &= \sigma \, 4\pi b^{2},\\ q_{c} &= \sigma \, 4\pi c^{2}. \end{aligned}$$

The total given charge is $$Q$$, so

$$Q \;=\; q_{a}+q_{b}+q_{c} \;=\; \sigma \, 4\pi \left(a^{2}+b^{2}+c^{2}\right).$$

Solving this expression for $$\sigma$$ gives

$$\sigma \;=\; \dfrac{Q}{4\pi\left(a^{2}+b^{2}+c^{2}\right)}.$$

Next we calculate the electric potential at a point whose distance from the common centre is $$r$$, with $$r<a$$. For a charged conducting spherical shell, the standard electrostatic result states:

$$\text{Potential at any interior point} \;=\; \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{q}{R},$$

because the entire shell’s charge may be treated as if it were concentrated at its centre for points lying inside the shell.

Hence the contributions of the three shells to the required potential $$V(r)$$ are

$$\begin{aligned} V_{a} &= \dfrac{1}{4\pi\varepsilon_{0}}\,\dfrac{q_{a}}{a},\\[4pt] V_{b} &= \dfrac{1}{4\pi\varepsilon_{0}}\,\dfrac{q_{b}}{b},\\[4pt] V_{c} &= \dfrac{1}{4\pi\varepsilon_{0}}\,\dfrac{q_{c}}{c}. \end{aligned}$$

Adding these, the total potential is

$$\begin{aligned} V(r) &= \dfrac{1}{4\pi\varepsilon_{0}}\left(\dfrac{q_{a}}{a}+\dfrac{q_{b}}{b}+\dfrac{q_{c}}{c}\right)\\[6pt] &= \dfrac{1}{4\pi\varepsilon_{0}}\left[\dfrac{\sigma\,4\pi a^{2}}{a}+\dfrac{\sigma\,4\pi b^{2}}{b}+\dfrac{\sigma\,4\pi c^{2}}{c}\right]\\[6pt] &= \dfrac{1}{4\pi\varepsilon_{0}}\;4\pi\sigma\left(a+b+c\right)\\[6pt] &= \dfrac{\sigma\left(a+b+c\right)}{\varepsilon_{0}}. \end{aligned}$$

Now we substitute the value of $$\sigma$$ obtained earlier:

$$\begin{aligned} V(r) &= \dfrac{1}{\varepsilon_{0}}\left[\dfrac{Q}{4\pi\left(a^{2}+b^{2}+c^{2}\right)}\right]\left(a+b+c\right)\\[8pt] &= \dfrac{Q\left(a+b+c\right)}{4\pi\varepsilon_{0}\left(a^{2}+b^{2}+c^{2}\right)}. \end{aligned}$$

This expression matches Option D.

Hence, the correct answer is Option D.