If `1/a,1/b,1/c` = 0 then `|(1+a,1,1),(1,1+b,1),(1,1,1+c)|` is equal to

Options

A. 0
B. abc
C. `-abc`
D. none of these

Answer

To solve the problem, we need to evaluate the determinant

\[ D = \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} \]

Given that \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \).

### Step 1: Expand the Determinant
We will expand the determinant along the first row (R1):

\[ D = (1+a) \begin{vmatrix} 1+b & 1 \\ 1 & 1+c \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & 1+c \end{vmatrix} + 1 \begin{vmatrix} 1 & 1+b \\ 1 & 1 \end{vmatrix} \]

### Step 2: Calculate the 2x2 Determinants
Now we calculate the 2x2 determinants:

1. For the first determinant:
\[ \begin{vmatrix} 1+b & 1 \\ 1 & 1+c \end{vmatrix} = (1+b)(1+c) - (1)(1) = (1+b+c+bc) - 1 = b + c + bc \]

2. For the second determinant:
\[ \begin{vmatrix} 1 & 1 \\ 1 & 1+c \end{vmatrix} = (1)(1+c) - (1)(1) = 1 + c - 1 = c \]

3. For the third determinant:
\[ \begin{vmatrix} 1 & 1+b \\ 1 & 1 \end{vmatrix} = (1)(1) - (1)(1+b) = 1 - (1+b) = -b \]

### Step 3: Substitute Back into the Determinant
Substituting these results back into the expression for \(D\):

\[ D = (1+a)(b+c+bc) - c - b \]

### Step 4: Expand and Simplify
Now we expand \(D\):

\[ D = (1+a)(b+c+bc) - c - b \]
\[ = (b+c+bc) + a(b+c+bc) - c - b \]
\[ = b + c + bc + ab + ac + abc - c - b \]
\[ = ab + ac + bc + abc \]

### Step 5: Factor Out abc
Now we can factor out \(abc\):

\[ D = abc \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 1 \right) \]

### Step 6: Use the Given Condition
Given that \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \):

\[ D = abc \left( 0 + 1 \right) = abc \]

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

\[ D = abc \]