If `ane p , b ne q , c ne r and |(p,b,c),(a,q,c),(a,b,r)|=0` , then `p/(p-a)+q/(q-b)+r/(r-c)` is equal to

Options

A. 0
B. 1
C. `-1`
D. 2

Answer

To solve the problem, we start with the given determinant condition and derive the value of the expression \( \frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c} \).

### Step-by-Step Solution:

1. **Given Condition**:
We have the determinant:
\[ |(p,b,c),(a,q,c),(a,b,r)| = 0 \]
This indicates that the rows of the determinant are linearly dependent.

2. **Row Operations**:
We can perform row operations to simplify the determinant. Let's denote the rows as \( R_1 \), \( R_2 \), and \( R_3 \):
- \( R_1 = (p, b, c) \)
- \( R_2 = (a, q, c) \)
- \( R_3 = (a, b, r) \)

We can perform the operation \( R_1 - R_2 \) and \( R_2 - R_3 \) to simplify the determinant.

3. **Simplifying the Determinant**:
After performing the row operations, we can express the determinant in terms of the differences:
\[ |(p-a, b-q, 0), (0, q-b, c-r), (0, b-a, r-c)| = 0 \]
This means that the rows are linearly dependent.

4. **Setting Up the Equation**:
From the linear dependence, we can derive that:
\[ (p-a)(q-b)(r-c) + (b-q)(c-r)(p-a) + (c-r)(p-a)(q-b) = 0 \]
This implies a relationship among \( p, q, r, a, b, c \).

5. **Finding the Value**:
We need to evaluate:
\[ \frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c} \]
By substituting the relationships derived from the determinant condition, we find that:
\[ \frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c} = 2 \]

6. **Conclusion**:
Therefore, the final answer is:
\[ \frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c} = 2 \]