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 problem, we need to evaluate the determinant given by:

\[ \Delta = \begin{vmatrix} a-1 & a & a+1 \\ b-1 & b & b+1 \\ c-1 & c & c+1 \end{vmatrix} \]

where \( a, b, c \) are even natural numbers.

### Step 1: Write down the determinant
We start with the determinant:

\[ \Delta = \begin{vmatrix} a-1 & a & a+1 \\ b-1 & b & b+1 \\ c-1 & c & c+1 \end{vmatrix} \]

### Step 2: Simplify the determinant using column operations
We can perform column operations to simplify the determinant. Let's subtract the second column from the first and the third column from the second:

\[ \Delta = \begin{vmatrix} (a-1) - a & a & (a+1) - a \\ (b-1) - b & b & (b+1) - b \\ (c-1) - c & c & (c+1) - c \end{vmatrix} \]

This simplifies to:

\[ \Delta = \begin{vmatrix} -1 & a & 1 \\ -1 & b & 1 \\ -1 & c & 1 \end{vmatrix} \]

### Step 3: Factor out -1 from the first column
We can factor out -1 from the first column:

\[ \Delta = -1 \cdot \begin{vmatrix} 1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1 \end{vmatrix} \]

### Step 4: Evaluate the determinant
Now we can evaluate the determinant:

\[ \Delta = -1 \cdot \begin{vmatrix} 1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1 \end{vmatrix} \]

Notice that the first column is the same for all rows, which means the determinant is zero:

\[ \Delta = 0 \]

### Conclusion
Since the value of the determinant \(\Delta\) is zero, it is indeed a multiple of any integer, including 4, 6, or 9.

### Final Answer
Thus, the answer is that \(\Delta\) is a multiple of 4, 6, and 9.

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