Let `A=[a_("ij")]_(3xx3), B=[b_("ij")]_(3xx3)` and `C=[c_("ij")]_(3xx3)` be any three matrices, where `b_("ij")=3^(i-j) a_("ij")` and `c_("ij")=4^(i-j) b_("ij")`. If det. `A=2`, then det. `(BC)` is equal to _______ .

Explanation

To solve the problem, we need to find the determinant of the product of matrices \( B \) and \( C \), given the relationships between the matrices and the determinant of matrix \( A \).

### Step-by-Step Solution:

1. **Understanding the matrices**:
- We have three matrices \( A \), \( B \), and \( C \) defined as follows:
- \( B_{ij} = 3^{i-j} A_{ij} \)
- \( C_{ij} = 4^{i-j} B_{ij} \)

2. **Finding the determinant of matrix \( B \)**:
- The determinant of \( B \) can be expressed in terms of the determinant of \( A \):
\[ \text{det}(B) = \text{det}(3^{i-j} A_{ij}) = 3^{\text{sum of indices}} \cdot \text{det}(A) \]
- For a \( 3 \times 3 \) matrix, the sum of the indices \( i-j \) for all elements will yield a factor of \( 3^0 \) for \( B_{11} \), \( 3^1 \) for \( B_{21} \), and so on. Thus, we can factor out \( 3^3 \) from the determinant:
\[ \text{det}(B) = 3^3 \cdot \text{det}(A) = 27 \cdot 2 = 54 \]

3. **Finding the determinant of matrix \( C \)**:
- Similarly, we find the determinant of \( C \):
\[ \text{det}(C) = \text{det}(4^{i-j} B_{ij}) = 4^{\text{sum of indices}} \cdot \text{det}(B) \]
- Again, for a \( 3 \times 3 \) matrix, the sum of the indices \( i-j \) will yield a factor of \( 4^0 \) for \( C_{11} \), \( 4^1 \) for \( C_{21} \), and so on. Thus, we can factor out \( 4^3 \) from the determinant:
\[ \text{det}(C) = 4^3 \cdot \text{det}(B) = 64 \cdot 54 = 3456 \]

4. **Finding the determinant of the product \( BC \)**:
- Using the property of determinants:
\[ \text{det}(BC) = \text{det}(B) \cdot \text{det}(C) \]
- Substituting the values we found:
\[ \text{det}(BC) = 54 \cdot 3456 = 186624 \]

### Final Answer:
\[ \text{det}(BC) = 186624 \]