`|{:(a-b,b-c,c-a),(x-y,y-z,z-x),(p-q,q-r,r-p):}|`

Options

A. abc
B. `p+q+r`
C. `x^2y^2z^2`
D. 0

Answer

To solve the determinant given by the matrix:

\[ D = \begin{vmatrix} a-b & b-c & c-a \\ x-y & y-z & z-x \\ p-q & q-r & r-p \end{vmatrix} \]

we can follow these steps:

### Step 1: Rewrite the Determinant
We start with the determinant as it is given:

\[ D = \begin{vmatrix} a-b & b-c & c-a \\ x-y & y-z & z-x \\ p-q & q-r & r-p \end{vmatrix} \]

### Step 2: Apply Column Operation
We can simplify the determinant by performing a column operation. Specifically, we can add all three columns together to the first column. This means we will replace the first column with the sum of all three columns:

\[ C_1 \rightarrow C_1 + C_2 + C_3 \]

This gives us:

\[ D = \begin{vmatrix} (a-b) + (b-c) + (c-a) & b-c & c-a \\ (x-y) + (y-z) + (z-x) & y-z & z-x \\ (p-q) + (q-r) + (r-p) & q-r & r-p \end{vmatrix} \]

### Step 3: Simplify Each Row
Now, we simplify each entry in the first column:

1. For the first row:
\((a-b) + (b-c) + (c-a) = a - b + b - c + c - a = 0\)

2. For the second row:
\((x-y) + (y-z) + (z-x) = x - y + y - z + z - x = 0\)

3. For the third row:
\((p-q) + (q-r) + (r-p) = p - q + q - r + r - p = 0\)

So, we have:

\[ D = \begin{vmatrix} 0 & b-c & c-a \\ 0 & y-z & z-x \\ 0 & q-r & r-p \end{vmatrix} \]

### Step 4: Identify the Determinant
Now, we can observe that the first column consists entirely of zeros.

### Step 5: Conclusion
The determinant of any matrix with a column (or row) of zeros is zero. Therefore, we conclude:

\[ D = 0 \]

### Final Answer
Thus, the value of the determinant is:

\[ \boxed{0} \]