If `a,b,c` are non-zero real number such that `|(bc,ca,ab),(ca,ab,bc),(ab,bc,ca)|=0,` then

Options

A. `(1)/(a) + (1)/(b) + (1)/(c) = 0`
B. `(1)/(a) - (1)/(b) - (1)/(c) = 0`
C. `(1)/(b) + (1)/(c) - (1)/(a) = 0`
D. `(1)/(b) - (1)/(c) - (1)/(a) = 0`

Answer

To solve the given problem, we need to evaluate the determinant of the matrix formed by the elements \( (bc, ca, ab), (ca, ab, bc), (ab, bc, ca) \) and set it equal to zero.

### Step-by-Step Solution:

1. **Write the Determinant**:
We need to compute the determinant:
\[ D = \begin{vmatrix} bc & ca & ab \\ ca & ab & bc \\ ab & bc & ca \end{vmatrix} \]

2. **Expand the Determinant**:
We can expand this determinant along the first row:
\[ D = bc \begin{vmatrix} ab & bc \\ bc & ca \end{vmatrix} - ca \begin{vmatrix} ca & bc \\ ab & ca \end{vmatrix} + ab \begin{vmatrix} ca & ab \\ ab & bc \end{vmatrix} \]

3. **Calculate the 2x2 Determinants**:
Now we compute the 2x2 determinants:
- For the first determinant:
\[ \begin{vmatrix} ab & bc \\ bc & ca \end{vmatrix} = ab \cdot ca - bc \cdot bc = abc - b^2c^2 \]
- For the second determinant:
\[ \begin{vmatrix} ca & bc \\ ab & ca \end{vmatrix} = ca \cdot ca - bc \cdot ab = c^2a^2 - abc \]
- For the third determinant:
\[ \begin{vmatrix} ca & ab \\ ab & bc \end{vmatrix} = ca \cdot bc - ab \cdot ab = cabc - a^2b^2 \]

4. **Substituting Back**:
Substitute these back into the determinant:
\[ D = bc(abc - b^2c^2) - ca(c^2a^2 - abc) + ab(cabc - a^2b^2) \]

5. **Simplifying**:
Expanding this gives:
\[ D = b^2c^2a - b^3c^3 - c^2a^3 + abc^2 + abc^2 - a^2b^2c \]
Collecting like terms, we can rewrite:
\[ D = a^3b + b^3c + c^3a - 3abc(a + b + c) \]

6. **Setting the Determinant to Zero**:
Since we know that \( D = 0 \), we have:
\[ a^3b + b^3c + c^3a - 3abc(a + b + c) = 0 \]

7. **Using the Identity**:
We can use the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
This implies that either \( a + b + c = 0 \) or \( a^2 + b^2 + c^2 - ab - ac - bc = 0 \).

8. **Conclusion**:
If \( a + b + c = 0 \), then:
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \]
Hence, the final result is:
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \]