`Delta=|{:(p,q,r),(p+2a,q+2b,r+2c),(a,b,c):}|` then

Options

A. a) `Delta=0`
B. b) `Delta=abc`
C. c) `Delta=pqr`
D. d) None of these

Answer

To solve the given problem, we need to evaluate the determinant:

\[ \Delta = \begin{vmatrix} p & q & r \\ p + 2a & q + 2b & r + 2c \\ a & b & c \end{vmatrix} \]

### Step 1: Write the determinant
We start with the determinant as given:

\[ \Delta = \begin{vmatrix} p & q & r \\ p + 2a & q + 2b & r + 2c \\ a & b & c \end{vmatrix} \]

### Step 2: Apply row operations
We will perform a row operation to simplify the determinant. Let's subtract the first row from the second row after multiplying the third row by 2. This means we will replace the second row \( R_2 \) with \( R_2 - R_1 - 2R_3 \).

\[ R_2 = R_2 - R_1 - 2R_3 \]

Calculating the new second row:

- First element: \( (p + 2a) - p - 2a = 0 \)
- Second element: \( (q + 2b) - q - 2b = 0 \)
- Third element: \( (r + 2c) - r - 2c = 0 \)

Thus, the determinant becomes:

\[ \Delta = \begin{vmatrix} p & q & r \\ 0 & 0 & 0 \\ a & b & c \end{vmatrix} \]

### Step 3: Evaluate the determinant
Now, we can see that the second row is entirely zeros. The property of determinants states that if any row (or column) of a determinant is all zeros, then the value of that determinant is zero.

Thus, we have:

\[ \Delta = 0 \]

### Conclusion
Therefore, we conclude that:

\[ \Delta = 0 \]

### Final Answer
The correct option is (a) \( \Delta = 0 \).

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