If `Delta=|(0,b-a,c-a),(a-b,0,c-b),(a-c,b-c,0)|`, then `Delta` eqauals

Options

A. `a + b + c`
B. `-(a + b +c)`
C. `abc`
D. 0

Explanation

To solve the determinant \( \Delta = \begin{vmatrix} 0 & b-a & c-a \\ a-b & 0 & c-b \\ a-c & b-c & 0 \end{vmatrix} \), we will expand the determinant along the first row.

### Step 1: Write the determinant
\[ \Delta = \begin{vmatrix} 0 & b-a & c-a \\ a-b & 0 & c-b \\ a-c & b-c & 0 \end{vmatrix} \]

### Step 2: Expand along the first row
Using the first row for expansion, we have:
\[ \Delta = 0 \cdot \begin{vmatrix} 0 & c-b \\ b-c & 0 \end{vmatrix} - (b-a) \cdot \begin{vmatrix} a-b & c-b \\ a-c & 0 \end{vmatrix} + (c-a) \cdot \begin{vmatrix} a-b & 0 \\ a-c & b-c \end{vmatrix} \]

### Step 3: Simplify the first term
The first term is zero because it is multiplied by 0:
\[ 0 \cdot \begin{vmatrix} 0 & c-b \\ b-c & 0 \end{vmatrix} = 0 \]

### Step 4: Calculate the second term
Now we need to calculate the second term:
\[ -(b-a) \cdot \begin{vmatrix} a-b & c-b \\ a-c & 0 \end{vmatrix} \]
Calculating the determinant:
\[ \begin{vmatrix} a-b & c-b \\ a-c & 0 \end{vmatrix} = (a-b) \cdot 0 - (c-b)(a-c) = -(c-b)(a-c) \]
Thus, the second term becomes:
\[ -(b-a)(-(c-b)(a-c)) = (b-a)(c-b)(a-c) \]

### Step 5: Calculate the third term
Now we calculate the third term:
\[ (c-a) \cdot \begin{vmatrix} a-b & 0 \\ a-c & b-c \end{vmatrix} \]
Calculating the determinant:
\[ \begin{vmatrix} a-b & 0 \\ a-c & b-c \end{vmatrix} = (a-b)(b-c) - 0 = (a-b)(b-c) \]
Thus, the third term becomes:
\[ (c-a)(a-b)(b-c) \]

### Step 6: Combine all terms
Now we combine all the terms:
\[ \Delta = 0 + (b-a)(c-b)(a-c) + (c-a)(a-b)(b-c) \]
Notice that both terms are symmetric in nature and can be rearranged.

### Step 7: Factor out common terms
We can see that both terms can be factored:
\[ \Delta = (b-a)(c-b)(a-c) + (c-a)(a-b)(b-c) = 0 \]

### Conclusion
Thus, we find that:
\[ \Delta = 0 \]