Step-by-step Solution:

1. Identify the objects and forces involved.

   • We have two solid spheres, both of radius R.
   • Sphere 1 (the 'light' sphere): density $\rho_1 = \rho$.
   • Sphere 2 (the 'heavy' sphere): density $\rho_2 = 3\rho$.
   • They are connected by a massless spring with force constant $k$.
   • The system is placed in a liquid with density $\rho_L = 2\rho$.
   • The forces acting on each sphere are its weight, the buoyant force from the liquid, and the spring force.

2. Calculate the weight and maximum buoyant force for each sphere.

   • The volume of each sphere is $V = \frac{4}{3}\pi R^3$.
   • Sphere 1 (light sphere):
     • Weight: $W_1 = m_1 g = (V \rho_1) g = (\frac{4}{3}\pi R^3 \rho) g$.
     • Maximum buoyant force (when fully submerged): $F_{B1,max} = V \rho_L g = (\frac{4}{3}\pi R^3)(2\rho) g = \frac{8}{3}\pi R^3 \rho g$.
   • Sphere 2 (heavy sphere):
     • Weight: $W_2 = m_2 g = (V \rho_2) g = (\frac{4}{3}\pi R^3)(3\rho) g = 4\pi R^3 \rho g$.
     • Maximum buoyant force (when fully submerged): $F_{B2,max} = V \rho_L g = (\frac{4}{3}\pi R^3)(2\rho) g = \frac{8}{3}\pi R^3 \rho g$.

3. Determine if the spheres are fully or partially submerged by analyzing the system as a whole.

   • The spring force is an internal force, so for the equilibrium of the entire system, we only consider the total weight and total buoyant force.
   • Total weight of the system: $W_{total} = W_1 + W_2 = \frac{4}{3}\pi R^3 \rho g + 4\pi R^3 \rho g = \frac{16}{3}\pi R^3 \rho g$.
   • Total maximum buoyant force (if both spheres are fully submerged): $F_{B,total,max} = F_{B1,max} + F_{B2,max} = \frac{8}{3}\pi R^3 \rho g + \frac{8}{3}\pi R^3 \rho g = \frac{16}{3}\pi R^3 \rho g$.
   • Since $W_{total} = F_{B,total,max}$, the system is in neutral equilibrium when both spheres are completely submerged. This means that the entire system can float at any depth as long as both spheres are fully immersed.

4. Evaluate options C and D.

   • From the analysis in step 3, we concluded that both spheres, including the light sphere (density $\rho$), must be completely submerged for the system to be in equilibrium.
   • Therefore, statement D is correct and statement C is incorrect.

5. Determine the elongation of the spring.

   • Since the light sphere ($ho_1 = \rho$) tends to float in the liquid ($ho_L = 2\rho$) and the heavy sphere ($ho_2 = 3\rho$) tends to sink, the spring connecting them will be stretched (elongated).
   • Let the elongation of the spring be $x$. The spring force is $F_s = kx$.
   • We can analyze the forces on either sphere at equilibrium. Let's use the heavy sphere (S2).

6. Free Body Diagram for the heavy sphere (S2).

   • Downward force: Weight $W_2$.
   • Upward forces: Buoyant force $F_{B2}$ and spring force $F_s$.
   • Since S2 is completely submerged, $F_{B2} = F_{B2,max}$.
   • At equilibrium: $W_2 = F_{B2} + F_s$.
   • Substituting the values: $4\pi R^3 \rho g = \frac{8}{3}\pi R^3 \rho g + kx$
   • Solving for $kx$: $kx = 4\pi R^3 \rho g - \frac{8}{3}\pi R^3 \rho g = \frac{12 - 8}{3}\pi R^3 \rho g = \frac{4}{3}\pi R^3 \rho g$
   • The net elongation of the spring is: $x = \frac{4\pi R^3 \rho g}{3k}$

7. (Verification) Free Body Diagram for the light sphere (S1).

   • Downward forces: Weight $W_1$ and spring force $F_s$ (pulling it down).
   • Upward force: Buoyant force $F_{B1}$.
   • Since S1 is completely submerged, $F_{B1} = F_{B1,max}$.
   • At equilibrium: $F_{B1} = W_1 + F_s$.
   • Substituting the values: $\frac{8}{3}\pi R^3 \rho g = \frac{4}{3}\pi R^3 \rho g + kx$
   • Solving for $kx$: $kx = \frac{8}{3}\pi R^3 \rho g - \frac{4}{3}\pi R^3 \rho g = \frac{4}{3}\pi R^3 \rho g$
   • This gives the same result for the spring force, confirming our calculation.

8. Evaluate options A and B.

   • Our calculated elongation is $x = \frac{4\pi R^3 \rho g}{3k}$.
   • Therefore, statement A is correct and statement B is incorrect.

Conclusion

The correct statements are A and D.