If a, b and c are non-zero real numbers and `|(1+a,1,1),(1,1+b,1),(1,1,1+c)|=0`, then what is the value of `(1)/(a) + (1)/(b) + (1)/(c)` ?

Options

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

Answer

To solve the problem, we need to evaluate the determinant and find the value of \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\) given that the determinant equals zero.

### Step 1: Write the determinant
We have the determinant:
\[ D = \begin{vmatrix} 1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c \end{vmatrix} \]

### Step 2: Apply row operations
We can simplify the determinant by applying the operation \(C_2 - C_1\) and \(C_3 - C_1\). This gives us:
\[ D = \begin{vmatrix} 1 + a & 0 & 0 \\ 1 & b & 0 \\ 1 & 0 & c \end{vmatrix} \]

### Step 3: Evaluate the determinant
Now we can evaluate the determinant:
\[ D = (1 + a) \begin{vmatrix} b & 0 \\ 0 & c \end{vmatrix} - 0 + 0 \]
Calculating the 2x2 determinant:
\[ D = (1 + a)(bc) \]

### Step 4: Set the determinant to zero
Since we are given that the determinant equals zero:
\[ (1 + a)(bc) = 0 \]
Since \(a\), \(b\), and \(c\) are non-zero, we conclude:
\[ 1 + a = 0 \implies a = -1 \]

### Step 5: Find the relationship for \(b\) and \(c\)
Similarly, we can apply the same reasoning for \(b\) and \(c\):
\[ 1 + b = 0 \implies b = -1 \]
\[ 1 + c = 0 \implies c = -1 \]

### Step 6: Calculate \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\)
Now we can find:
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{-1} + \frac{1}{-1} + \frac{1}{-1} = -1 - 1 - 1 = -3 \]

### Final Answer
Thus, the value of \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\) is:
\[ \boxed{-3} \]