If `1 + (1)/(a) + (1)/(b) + (1)/(c) = 0`, then
`Delta = |(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 \( \Delta = |(1 + a, 1, 1), (1, 1 + b, 1), (1, 1, 1 + c)| \) given the condition \( 1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \).

### Step 1: Write the Determinant
We start with the determinant:

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

### Step 2: Simplify the Determinant
We can perform column operations to simplify the determinant. Let's change the second column by subtracting the third column from it:

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

This simplifies to:

\[ \Delta = \begin{vmatrix} 1 + a & 0 & 1 \\ 1 & b & 1 \\ 1 & -c & 1 + c \end{vmatrix} \]

### Step 3: Further Simplification
Next, we can perform a row operation by subtracting the first row from the second and third rows:

\[ \Delta = \begin{vmatrix} 1 + a & 0 & 1 \\ 1 - (1 + a) & b & 1 - 1 \\ 1 - (1 + a) & -c & (1 + c) - 1 \end{vmatrix} \]

This simplifies to:

\[ \Delta = \begin{vmatrix} 1 + a & 0 & 1 \\ -a & b & 0 \\ -a & -c & c \end{vmatrix} \]

### Step 4: Calculate the Determinant
Now we can calculate the determinant using the formula for a \(3 \times 3\) determinant:

\[ \Delta = (1 + a) \begin{vmatrix} b & 0 \\ -c & c \end{vmatrix} - 0 + 1 \begin{vmatrix} -a & b \\ -a & -c \end{vmatrix} \]

Calculating the first determinant:

\[ \begin{vmatrix} b & 0 \\ -c & c \end{vmatrix} = bc \]

Calculating the second determinant:

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

Thus, we have:

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

### Step 5: Substitute the Given Condition
Now, we substitute the condition \(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0\) which implies:

\[ 1 + \frac{bc + ac + ab}{abc} = 0 \implies bc + ac + ab = -abc \]

Substituting this back into our expression for \(\Delta\):

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

### Step 6: Final Calculation
Combining terms, we can express \(\Delta\) in terms of \(abc\):

\[ \Delta = abc \cdot 0 = 0 \]

Thus, the value of the determinant \( \Delta \) is:

\[ \Delta = 0 \]