Given `a_(i)^(2) + b_(i)^(2) + c_(i)^(2) = 1, i = 1, 2, 3 and a_(i) a_(j) + b_(i) b_(j) + c_(i) c_(j) = 0 (i !=j, i, j =1, 2, 3)`, then the value of the determinant
`|(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3))|`, is

Options

A. `(1)/(2)`
B. 0
C. 2
D. 1

Explanation

To solve the problem, we need to evaluate the determinant given the conditions on the vectors defined by \( a_i, b_i, c_i \).

### Step-by-Step Solution:

1. **Understanding the Conditions**:
We have three conditions:
- \( a_i^2 + b_i^2 + c_i^2 = 1 \) for \( i = 1, 2, 3 \)
- \( a_i a_j + b_i b_j + c_i c_j = 0 \) for \( i \neq j \)

These conditions imply that the vectors \( \mathbf{P} = (a_1, b_1, c_1) \), \( \mathbf{Q} = (a_2, b_2, c_2) \), and \( \mathbf{R} = (a_3, b_3, c_3) \) are unit vectors (length 1) and are mutually orthogonal.

2. **Forming the Determinant**:
We need to evaluate the determinant:
\[ D = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \]

3. **Properties of Orthogonal Vectors**:
Since the vectors \( \mathbf{P}, \mathbf{Q}, \mathbf{R} \) are orthogonal and of unit length, the determinant of the matrix formed by these vectors as columns will be equal to the volume of the parallelepiped formed by these vectors.

4. **Calculating the Determinant**:
The volume of the parallelepiped formed by three orthogonal unit vectors is given by:
\[ \text{Volume} = |\mathbf{P} \cdot (\mathbf{Q} \times \mathbf{R})| = 1 \]
Hence, the determinant \( D \) is equal to 1.

5. **Conclusion**:
Therefore, the value of the determinant is:
\[ D = 1 \]