In the following equilibria
`I:A+2B hArr C, K_(eq)=K_(1)`
`II: C+DhArr 3A, K_(eq)=K_(2)`
`III, 6B+D hArr 2C, K_(eq)=K_(3)`
Hence,

Options

A. `3K_(1)+K_(2)=K_(3)`
B. `K_(1)^(3).K_(2)^(2)=K_(3)`
C. `3K_(1)+K_(2)^(2)=K_(3)`
D. `K_(1)^(3).K_(2)=K_(3)`

Explanation

To solve the problem, we need to analyze the three given equilibrium reactions and their constants. We will derive the relationship between the equilibrium constants \( K_1 \), \( K_2 \), and \( K_3 \) based on the reactions provided.

### Step-by-Step Solution:

1. **Identify the Reactions and Their Equilibrium Constants:**
- Reaction I: \( A + 2B \rightleftharpoons C \) with \( K_{eq} = K_1 \)
- Reaction II: \( C + D \rightleftharpoons 3A \) with \( K_{eq} = K_2 \)
- Reaction III: \( 6B + D \rightleftharpoons 2C \) with \( K_{eq} = K_3 \)

2. **Manipulate Reaction I:**
- Multiply Reaction I by 3:
\[ 3(A + 2B \rightleftharpoons C) \implies 3A + 6B \rightleftharpoons 3C \]
- The equilibrium constant for this modified reaction becomes:
\[ K_1^3 \]

3. **Combine the Modified Reaction I with Reaction II:**
- We have the modified Reaction I:
\[ 3A + 6B \rightleftharpoons 3C \quad (K_1^3) \]
- And Reaction II:
\[ C + D \rightleftharpoons 3A \quad (K_2) \]
- To combine, we can reverse Reaction II:
\[ 3A \rightleftharpoons C + D \quad (K_2^{-1}) \]

4. **Add the Two Reactions:**
- Adding the modified Reaction I and the reversed Reaction II:
\[ (3A + 6B \rightleftharpoons 3C) + (3A \rightleftharpoons C + D) \]
- This results in:
\[ 6B + D \rightleftharpoons 2C \]
- The equilibrium constant for this combined reaction is:
\[ K_3 = K_1^3 \cdot K_2^{-1} \]

5. **Rearranging the Equation:**
- To express \( K_3 \) in terms of \( K_1 \) and \( K_2 \):
\[ K_3 = \frac{K_1^3}{K_2} \]

### Conclusion:
The relationship between the equilibrium constants is:
\[ K_3 = \frac{K_1^3}{K_2} \]