If a, b and c are all different from zero and
`Delta = |(1 +a,1,1),(1,1 +b,1),(1,1,1 +c)| = 0`, then the value of `(1)/(a) + (1)/(b) + (1)/(c)` is

Options

A. abc
B. `(1)/(abc)`
C. `-a -b -c`
D. `-1`

Answer

To solve the problem, we need to evaluate the expression \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\) given that the determinant \(\Delta = |(1 + a, 1, 1), (1, 1 + b, 1), (1, 1, 1 + c)| = 0\).

### Step-by-Step Solution:

1. **Set Up the Determinant**:
\[ \Delta = \begin{vmatrix} 1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c \end{vmatrix} \]

2. **Expand the Determinant**:
We can use the formula for the determinant of a \(3 \times 3\) matrix:
\[ \Delta = (1 + a)\begin{vmatrix} 1 + b & 1 \\ 1 & 1 + c \end{vmatrix} - 1\begin{vmatrix} 1 & 1 \\ 1 & 1 + c \end{vmatrix} + 1\begin{vmatrix} 1 & 1 + b \\ 1 & 1 \end{vmatrix} \]

3. **Calculate the 2x2 Determinants**:
- First determinant:
\[ \begin{vmatrix} 1 + b & 1 \\ 1 & 1 + c \end{vmatrix} = (1 + b)(1 + c) - 1 \cdot 1 = 1 + b + c + bc - 1 = b + c + bc \]
- Second determinant:
\[ \begin{vmatrix} 1 & 1 \\ 1 & 1 + c \end{vmatrix} = 1(1 + c) - 1 \cdot 1 = 1 + c - 1 = c \]
- Third determinant:
\[ \begin{vmatrix} 1 & 1 + b \\ 1 & 1 \end{vmatrix} = 1 \cdot 1 - 1 \cdot (1 + b) = 1 - (1 + b) = -b \]

4. **Substituting Back**:
Now substituting these back into the determinant expression:
\[ \Delta = (1 + a)(b + c + bc) - c - b \]
Expanding this gives:
\[ \Delta = (1 + a)(b + c + bc) - c - b = b + c + bc + ab + ac + abc - c - b \]
Simplifying:
\[ \Delta = ab + ac + abc \]

5. **Setting the Determinant to Zero**:
Since we know \(\Delta = 0\):
\[ ab + ac + abc = 0 \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ abc = - (ab + ac) \]

7. **Finding the Value of \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\)**:
We can express \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\) as:
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{bc + ac + ab}{abc} \]
Substituting \(abc = - (ab + ac)\):
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{-(ab + ac)}{-abc} = \frac{ab + ac}{abc} = -1 \]

### Final Answer:
Thus, the value of \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\) is \(-1\).