If a,b,c are non zero real numbers then
`Delta=|(1,ab,1/a+1/b),(1,bc,1/b+1/c),(1,ca,1/c+1/a)|` equals

Options

A. 0
B. `bc+ca+ab`
C. `a^(-1)+b^(-1)+c^(-1)`
D. `abc-1`

Answer

To solve the determinant \( \Delta = \begin{vmatrix} 1 & ab & \frac{1}{a} + \frac{1}{b} \\ 1 & bc & \frac{1}{b} + \frac{1}{c} \\ 1 & ca & \frac{1}{c} + \frac{1}{a} \end{vmatrix} \), we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant as given:
\[ \Delta = \begin{vmatrix} 1 & ab & \frac{1}{a} + \frac{1}{b} \\ 1 & bc & \frac{1}{b} + \frac{1}{c} \\ 1 & ca & \frac{1}{c} + \frac{1}{a} \end{vmatrix} \]

### Step 2: Apply Row Operations
We can simplify the determinant by performing row operations. We will subtract the first row from the second and third rows:
\[ \Delta = \begin{vmatrix} 1 & ab & \frac{1}{a} + \frac{1}{b} \\ 0 & bc - ab & \left(\frac{1}{b} + \frac{1}{c}\right) - \left(\frac{1}{a} + \frac{1}{b}\right) \\ 0 & ca - ab & \left(\frac{1}{c} + \frac{1}{a}\right) - \left(\frac{1}{a} + \frac{1}{b}\right) \end{vmatrix} \]

### Step 3: Simplify the Second and Third Rows
Now, simplifying the second and third rows:
- For the second row:
\[ \left(\frac{1}{b} + \frac{1}{c}\right) - \left(\frac{1}{a} + \frac{1}{b}\right) = \frac{1}{c} - \frac{1}{a} = \frac{a - c}{ac} \]
- For the third row:
\[ \left(\frac{1}{c} + \frac{1}{a}\right) - \left(\frac{1}{a} + \frac{1}{b}\right) = \frac{1}{c} - \frac{1}{b} = \frac{b - c}{bc} \]

Thus, we have:
\[ \Delta = \begin{vmatrix} 1 & ab & \frac{1}{a} + \frac{1}{b} \\ 0 & bc - ab & \frac{a - c}{ac} \\ 0 & ca - ab & \frac{b - c}{bc} \end{vmatrix} \]

### Step 4: Factor Out Common Terms
Next, we can factor out \( (bc - ab) \) and \( (ca - ab) \) from the second and third rows respectively:
\[ \Delta = (bc - ab)(ca - ab) \begin{vmatrix} 1 & ab & \frac{1}{a} + \frac{1}{b} \\ 0 & 1 & \frac{a - c}{(bc - ab)ac} \\ 0 & 1 & \frac{b - c}{(ca - ab)bc} \end{vmatrix} \]

### Step 5: Observe the Columns
Notice that the second and third columns of the determinant are now identical (after appropriate simplifications). Therefore, the determinant evaluates to zero:
\[ \Delta = 0 \]

### Conclusion
Thus, the value of the determinant \( \Delta \) is:
\[ \Delta = 0 \]