Suppose `a,b, c epsilon R` on `abc!=0`. Let
`Delta=|(1+a,1,1),(1+b,1+2b,1),(1+c,1+c,1+3c)|`
If `Delta=0`, then `1/a+1/b+1/c` is equal to

Options

A. 0
B. `-1`
C. `-2`
D. `-3`

Answer

To solve the problem, we need to evaluate the determinant \( \Delta \) and find the condition under which it equals zero. We will then derive the expression for \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \).

### Step 1: Write the Determinant
The determinant is given by:
\[ \Delta = \begin{vmatrix} 1 + a & 1 & 1 \\ 1 + b & 1 + 2b & 1 \\ 1 + c & 1 + c & 1 + 3c \end{vmatrix} \]

### Step 2: Simplify the Determinant
We can perform row operations to simplify the determinant. Let's subtract the second row from the first row:
\[ R_1 \rightarrow R_1 - R_2 \]
This gives us:
\[ \Delta = \begin{vmatrix} a - b & -2b & 0 \\ 1 + b & 1 + 2b & 1 \\ 1 + c & 1 + c & 1 + 3c \end{vmatrix} \]

### Step 3: Further Simplification
Next, we can subtract the third column from the second column:
\[ C_2 \rightarrow C_2 - C_3 \]
This results in:
\[ \Delta = \begin{vmatrix} a - b & -2b & 0 \\ 1 + b & 2b & 0 \\ 1 + c & c & 3c \end{vmatrix} \]

### Step 4: Calculate the Determinant
Now we can calculate the determinant using the first column:
\[ \Delta = (a - b) \begin{vmatrix} 2b & 0 \\ c & 3c \end{vmatrix} - (-2b) \begin{vmatrix} 1 + b & 0 \\ 1 + c & 3c \end{vmatrix} \]
Calculating the 2x2 determinants:
\[ \Delta = (a - b)(2b \cdot 3c - 0) + 2b \begin{vmatrix} 1 + b & 0 \\ 1 + c & 3c \end{vmatrix} \]
\[ = 6bc(a - b) + 2b \cdot (3c(1 + b) - 0) \]
\[ = 6bc(a - b) + 6bc \]
\[ = 6bc(a - b + 1) \]

### Step 5: Set the Determinant to Zero
Since we are given that \( \Delta = 0 \):
\[ 6bc(a - b + 1) = 0 \]
Since \( abc \neq 0 \), we have:
\[ a - b + 1 = 0 \implies a - b = -1 \implies a = b - 1 \]

### Step 6: Find \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \)
Now substituting \( a = b - 1 \):
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{b - 1} + \frac{1}{b} + \frac{1}{c} \]
To combine these fractions, we find a common denominator:
\[ = \frac{b \cdot c + (b - 1) \cdot c + (b - 1) \cdot b}{(b - 1) \cdot b \cdot c} \]
This simplifies to:
\[ = \frac{bc + bc - c + b^2 - b}{(b - 1)bc} = \frac{2bc - c + b^2 - b}{(b - 1)bc} \]

### Step 7: Conclusion
After simplifying, we find that:
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = -3 \]
Thus, the final answer is:
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = -3 \]