If `x in R,a_(i),b_(i),c_(i) in R` for `i=1,2,3` and `|{:(a_(1)+b_(1)x,a_(1)x+b_(1),c_(1)),(a_(2)+b_(2)x,a_(2)x+b_(2),c_(2)),(a_(3)+b_(3)x,a_(3)x+b_(3),c_(3)):}|=0`, then which of the following may be true ?

Options

A. `x=1`
B. `x=-1`
C. `|{:(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3)):}|=0`
D. none of these

Explanation

To solve the given problem, we need to analyze the determinant provided and derive the conditions under which it equals zero. Let's break down the steps systematically.

### Step 1: Write down the determinant
We start with the determinant given in the problem:

\[ D = \begin{vmatrix} a_1 + b_1 x & a_1 x + b_1 & c_1 \\ a_2 + b_2 x & a_2 x + b_2 & c_2 \\ a_3 + b_3 x & a_3 x + b_3 & c_3 \end{vmatrix} \]

### Step 2: Expand the determinant
We can expand this determinant using the properties of determinants. We will keep the first column as it is and perform operations on the second and third columns.

### Step 3: Perform column operations
We can manipulate the second column by subtracting \(x\) times the first column from it:

\[ D = \begin{vmatrix} a_1 + b_1 x & (a_1 x + b_1) - x(a_1 + b_1 x) & c_1 \\ a_2 + b_2 x & (a_2 x + b_2) - x(a_2 + b_2 x) & c_2 \\ a_3 + b_3 x & (a_3 x + b_3) - x(a_3 + b_3 x) & c_3 \end{vmatrix} \]

This simplifies the second column.

### Step 4: Factor out common terms
After simplifying, we can factor out \(x\) from the second column, leading to a new determinant structure.

### Step 5: Analyze the resulting determinant
The determinant can be expressed as:

\[ D = \begin{vmatrix} a_1 + b_1 x & b_1 & c_1 \\ a_2 + b_2 x & b_2 & c_2 \\ a_3 + b_3 x & b_3 & c_3 \end{vmatrix} \]

### Step 6: Set the determinant equal to zero
Since we are given that the determinant equals zero, we have:

\[ D = 0 \]

### Step 7: Identify conditions for the determinant to be zero
For the determinant to be zero, either:

1. The rows (or columns) of the determinant must be linearly dependent.
2. The determinant can be expressed as a product of factors that equal zero.

From the determinant, we can derive that:

\[ 1 - x^2 = 0 \quad \text{or} \quad a_1 b_1 c_1 a_2 b_2 c_2 a_3 b_3 = 0 \]

### Step 8: Solve for \(x\)
From \(1 - x^2 = 0\), we find:

\[ x^2 = 1 \implies x = \pm 1 \]

### Step 9: Conclusion
Thus, the possible values of \(x\) are \(1\) and \(-1\). Additionally, the condition \(a_1 b_1 c_1 a_2 b_2 c_2 a_3 b_3 = 0\) must also hold true.

### Final Answer
The options that may be true are:
- \(x = 1\)
- \(x = -1\)
- The determinant \(a_1 b_1 c_1 a_2 b_2 c_2 a_3 b_3 = 0\)