The value of `|{:(a+pd,a+qd,a+rd),(p,q,r),(d,d,d):}|` is

Options

A. `-1`
B. 0
C. 1
D. `p+q+r`

Answer

To find the value of the determinant \( | \begin{vmatrix} a + pd & a + qd & a + rd \\ p & q & r \\ d & d & d \end{vmatrix} | \), we can follow these steps:

### Step 1: Rewrite the Determinant
Let:
\[ D = \begin{vmatrix} a + pd & a + qd & a + rd \\ p & q & r \\ d & d & d \end{vmatrix} \]

### Step 2: Factor Out Common Terms
Notice that in the first row, we can factor out \( a \) from each of the first row's elements:
\[ D = \begin{vmatrix} a + pd & a + qd & a + rd \\ p & q & r \\ d & d & d \end{vmatrix} = \begin{vmatrix} a & a & a \\ p & q & r \\ d & d & d \end{vmatrix} + \begin{vmatrix} pd & qd & rd \\ p & q & r \\ d & d & d \end{vmatrix} \]

### Step 3: Apply Properties of Determinants
Using the property of determinants, we can separate the determinant into two parts:
\[ D = \begin{vmatrix} a & a & a \\ p & q & r \\ d & d & d \end{vmatrix} + \begin{vmatrix} pd & qd & rd \\ p & q & r \\ d & d & d \end{vmatrix} \]

### Step 4: Analyze the Determinants
1. In the first determinant, the first row is \( (a, a, a) \) which means all elements are the same. Hence, the determinant is zero:
\[ \begin{vmatrix} a & a & a \\ p & q & r \\ d & d & d \end{vmatrix} = 0 \]

2. In the second determinant, the second row is \( (p, q, r) \) and the third row is \( (d, d, d) \), which also means that two rows are identical (the second and third rows are not identical here, but we will analyze it further). The first row is \( (pd, qd, rd) \). Since the third row is \( (d, d, d) \), the determinant will also evaluate to zero:
\[ \begin{vmatrix} pd & qd & rd \\ p & q & r \\ d & d & d \end{vmatrix} = 0 \]

### Step 5: Combine the Results
Since both determinants are zero, we can conclude:
\[ D = 0 + 0 = 0 \]

### Final Answer
Thus, the value of the determinant is:
\[ \boxed{0} \]