There are three co-centric conducting spherical shells A, B and C of radii a, b and c respectively (c > b > a) and they are charged with charge $$q_{1},q_{2}\text{ and }q_{3}$$ respectively. The potentials of the spheres A, Band C respectively, are :

We need to find the potentials of three concentric conducting spherical shells A, B, and C with radii $$a$$, $$b$$, and $$c$$ (where $$c > b > a$$) carrying charges $$q_1$$, $$q_2$$, and $$q_3$$ respectively.

Since the key principle is that the potential at any point due to a conducting spherical shell of radius R carrying charge q is $$\frac{q}{4\pi\epsilon_0 R}$$ when the point is on the shell or inside it ($$r \le R$$), and $$\frac{q}{4\pi\epsilon_0 r}$$ when the point is outside ($$r > R$$).

For sphere A at radius $$a$$, the contribution from $$q_1$$ on A itself is $$V_1 = \frac{q_1}{4\pi\epsilon_0 a}$$ since the point lies on A, from $$q_2$$ on B is $$V_2 = \frac{q_2}{4\pi\epsilon_0 b}$$ as the point is inside B, and from $$q_3$$ on C is $$V_3 = \frac{q_3}{4\pi\epsilon_0 c}$$ because the point is inside C.

Therefore, the total potential at A is $$V_A = \frac{1}{4\pi\epsilon_0}\left(\frac{q_1}{a} + \frac{q_2}{b} + \frac{q_3}{c}\right)$$.

Turning to sphere B at radius $$b$$, the potential due to $$q_1$$ on A is $$\frac{q_1}{4\pi\epsilon_0 b}$$ since B lies outside A, the potential from $$q_2$$ on B itself is $$\frac{q_2}{4\pi\epsilon_0 b}$$, and the potential from $$q_3$$ on C is $$\frac{q_3}{4\pi\epsilon_0 c}$$ because B is inside C.

Hence, the total potential at B is $$V_B = \frac{1}{4\pi\epsilon_0}\left(\frac{q_1 + q_2}{b} + \frac{q_3}{c}\right)$$.

Finally, for sphere C at radius $$c$$, both A and B lie inside C, so the contributions from $$q_1$$ and $$q_2$$ are $$\frac{q_1}{4\pi\epsilon_0 c}$$ and $$\frac{q_2}{4\pi\epsilon_0 c}$$ respectively, and the self-potential from $$q_3$$ is $$\frac{q_3}{4\pi\epsilon_0 c}$$.

Thus, the total potential at C is $$V_C = \frac{1}{4\pi\epsilon_0}\left(\frac{q_1 + q_2 + q_3}{c}\right)$$.

Comparing these results with the given options shows that they correspond to Option 2.