If a, b, c are even natural numbers, then `Delta=|{:(a-1,a,a+1),(b-1,b,b+1),(c-1,c,c+1):}|` is a multiple of

Options

A. 4
B. 6
C. 9
D. None of these

Answer

To solve the given problem, we need to evaluate the determinant \( \Delta = \begin{vmatrix} a-1 & b-1 & c-1 \\ a & b & c \\ a+1 & b+1 & c+1 \end{vmatrix} \) and show that it is a multiple of a certain number.

### Step-by-Step Solution:

1. **Set Up the Determinant**:
We start with the determinant:
\[ \Delta = \begin{vmatrix} a-1 & b-1 & c-1 \\ a & b & c \\ a+1 & b+1 & c+1 \end{vmatrix} \]

2. **Apply Row Operations**:
We can perform row operations to simplify the determinant. Let's subtract the second row from the first row and the second row from the third row:
\[ R_1 \rightarrow R_1 - R_2 \quad \text{and} \quad R_3 \rightarrow R_3 - R_2 \]
This gives us:
\[ \Delta = \begin{vmatrix} (a-1)-a & (b-1)-b & (c-1)-c \\ a & b & c \\ (a+1)-a & (b+1)-b & (c+1)-c \end{vmatrix} \]
Simplifying this, we have:
\[ \Delta = \begin{vmatrix} -1 & -1 & -1 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} \]

3. **Factor Out Common Terms**:
We can factor out \(-1\) from the first row:
\[ \Delta = -1 \cdot \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} \]

4. **Evaluate the Determinant**:
Notice that the first and third rows are identical. Therefore, the determinant evaluates to zero:
\[ \Delta = -1 \cdot 0 = 0 \]

5. **Conclusion**:
Since the value of the determinant \( \Delta \) is zero, it is a multiple of every integer.

### Final Answer:
Thus, \( \Delta \) is a multiple of every integer.