Let the common surface charge density be $\sigma$.

Since charge is distributed on three concentric spherical shells of radii $a, b, c$ with equal surface charge density, the charges on the shells are:

$q_a = 4\pi a^2 \sigma, \quad q_b = 4\pi b^2 \sigma, \quad q_c = 4\pi c^2 \sigma$

Use total charge $Q$.

$Q = q_a + q_b + q_c = 4\pi \sigma (a^2 + b^2 + c^2)$

Hence,

$\sigma = \frac{Q}{4\pi (a^2+b^2+c^2)}$

Potential at a point inside a spherical shell.

For a spherical shell of radius $R$ carrying charge $q$, the potential at any interior point is constant:

$V = \frac{1}{4\pi \varepsilon_0}\frac{q}{R}$

Since $r<a$, the point lies inside all three shells.

Therefore total potential is:

$V = \frac{1}{4\pi \varepsilon_0}\left(\frac{q_a}{a} + \frac{q_b}{b} + \frac{q_c}{c}\right)$

Substitute $q_a, q_b, q_c$.

$V = \frac{1}{4\pi \varepsilon_0}\left(\frac{4\pi a^2\sigma}{a} + \frac{4\pi b^2\sigma}{b} + \frac{4\pi c^2\sigma}{c}\right)$

$V = \frac{1}{4\pi \varepsilon_0} \cdot 4\pi \sigma (a+b+c)$

$V = \frac{\sigma (a+b+c)}{\varepsilon_0}$

Now substitute $\sigma$.

$V = \frac{1}{\varepsilon_0} \cdot \frac{Q}{4\pi (a^2+b^2+c^2)}(a+b+c)$

$V = \frac{Q(a+b+c)}{4\pi \varepsilon_0 (a^2+b^2+c^2)}$

Compare with options.

This matches Option D:

$\boxed{\frac{Q(a+b+c)}{4\pi \varepsilon_0 (a^2+b^2+c^2)}}$