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 omega) + (1)/(c omega^(2)) = 0`
B. `(1)/(a) + (1)/(b omega^(2)) + (1)/(c omega) = 0`
C. `(1)/(a omega) + (1)/(b omega^(2)) + (1)/(c) = 0`
D. `(1)/(a) + (1)/(b) + (1)/(c) = 0`

Answer

To solve the problem, we need to evaluate the determinant given by:

\[ D = \begin{vmatrix} bc & ca & ab \\ ca & ab & bc \\ ab & bc & ca \end{vmatrix} \]

We are given that \( |D| = 0 \).

### Step 1: Calculate the Determinant

Using the determinant formula for a \(3 \times 3\) matrix, we have:

\[ D = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \]

Substituting the values from our matrix:

\[ D = bc(ab \cdot ca - bc \cdot ab) - ca(ca \cdot ca - bc \cdot ab) + ab(ca \cdot bc - ab \cdot ab) \]

### Step 2: Simplify Each Term

1. The first term:
\[ bc(ab \cdot ca - bc \cdot ab) = bc(abca - bcab) = bc(abca - abbc) = 0 \]

2. The second term:
\[ -ca(ca^2 - b^2c^2) = -ca(c^2a^2 - b^2c^2) = -ca(c^2a^2 - b^2c^2) \]

3. The third term:
\[ ab(ca \cdot bc - ab^2) = ab(cabc - a^2b) = ab(cabc - a^2b) \]

### Step 3: Combine the Terms

Combining all the terms, we have:

\[ D = 0 - ca(c^2a^2 - b^2c^2) + ab(cabc - a^2b) \]

### Step 4: Set the Determinant to Zero

Since we know \(D = 0\), we can set the equation to zero:

\[ -ca(c^2a^2 - b^2c^2) + ab(cabc - a^2b) = 0 \]

### Step 5: Factor the Expression

Factoring out common terms, we can analyze the conditions under which this determinant equals zero.

### Step 6: Analyze Conditions

We can derive two conditions from the determinant being zero:

1. \(bc + ca + ab = 0\)
2. \(ab = bc = ca\)

From these conditions, we can deduce relationships between \(a\), \(b\), and \(c\).

### Step 7: Conclusion

From the conditions derived, we can conclude that if \(a\), \(b\), and \(c\) are non-zero, then they must be equal or related in such a way that their products yield a zero sum.