`Delta` is a 3rd order determinant having each element in `R_1` is sum of 2 terms and each element in `R_2` is sum of three terms and each element in 3rd row is sum of four terms. If `Delta=nDelta_1` , where `Delta_1` , is a determinant having single elements in each row, then n =

Options

A. 6
B. 12
C. 24
D. none

Answer

To solve the problem, we need to analyze the structure of the determinant \(\Delta\) based on the given information about the rows and then relate it to the simpler determinant \(\Delta_1\).

### Step-by-Step Solution:

1. **Understanding the Structure of the Determinant**:
- The determinant \(\Delta\) is a 3rd order determinant with three rows \(R_1\), \(R_2\), and \(R_3\).
- Each element in \(R_1\) is a sum of 2 terms, each element in \(R_2\) is a sum of 3 terms, and each element in \(R_3\) is a sum of 4 terms.

2. **Expressing the Rows**:
- Let’s denote the elements of the first row \(R_1\) as:
\[ R_1 = [a_1 + b_1, a_2 + b_2, a_3 + b_3] \]
- The second row \(R_2\) can be expressed as:
\[ R_2 = [c_1 + d_1 + e_1, c_2 + d_2 + e_2, c_3 + d_3 + e_3] \]
- The third row \(R_3\) can be expressed as:
\[ R_3 = [f_1 + g_1 + h_1 + i_1, f_2 + g_2 + h_2 + i_2, f_3 + g_3 + h_3 + i_3] \]

3. **Setting Up the Determinant**:
- The determinant \(\Delta\) can be written as:
\[ \Delta = \begin{vmatrix} a_1 + b_1 & c_1 + d_1 + e_1 & f_1 + g_1 + h_1 + i_1 \\ a_2 + b_2 & c_2 + d_2 + e_2 & f_2 + g_2 + h_2 + i_2 \\ a_3 + b_3 & c_3 + d_3 + e_3 & f_3 + g_3 + h_3 + i_3 \end{vmatrix} \]

4. **Expanding the Determinant**:
- We can expand this determinant using the linearity property. Each row can be expressed as a sum of terms:
\[ \Delta = \begin{vmatrix} a_1 & c_1 & f_1 \\ a_2 & c_2 & f_2 \\ a_3 & c_3 & f_3 \end{vmatrix} + \text{(other terms)} \]
- The other terms will arise from the combinations of the additional terms in each row.

5. **Relating to \(\Delta_1\)**:
- The determinant \(\Delta_1\) is defined as the determinant with single elements in each row. Thus, it can be expressed as:
\[ \Delta_1 = \begin{vmatrix} a_1 & c_1 & f_1 \\ a_2 & c_2 & f_2 \\ a_3 & c_3 & f_3 \end{vmatrix} \]
- The contributions from the additional terms in \(\Delta\) will scale the determinant \(\Delta_1\).

6. **Calculating the Scaling Factor \(n\)**:
- Each row contributes a factor based on the number of terms:
- The first row contributes a factor of 2 (since each element is a sum of 2 terms).
- The second row contributes a factor of 3 (since each element is a sum of 3 terms).
- The third row contributes a factor of 4 (since each element is a sum of 4 terms).
- Therefore, the total scaling factor \(n\) can be calculated as:
\[ n = 2 \times 3 \times 4 = 24 \]

### Final Answer:
Thus, \(n = 24\).