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 given problem, we need to evaluate the determinant

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

given the conditions:

1. \( a_i^2 + b_i^2 + c_i^2 = 1 \) for \( i = 1, 2, 3 \)
2. \( a_i a_j + b_i b_j + c_i c_j = 0 \) for \( i \neq j \)

### Step 1: Understanding the Conditions
The first condition states that each vector \( (a_i, b_i, c_i) \) has a magnitude of 1. This means that each vector lies on the unit sphere in 3D space.

### Step 2: Orthogonality of Vectors
The second condition indicates that the vectors \( (a_i, b_i, c_i) \) are mutually orthogonal. Specifically, for any two different indices \( i \) and \( j \), the dot product of the vectors is zero:

\[ (a_i, b_i, c_i) \cdot (a_j, b_j, c_j) = 0 \]

This implies that the vectors are orthogonal to each other.

### Step 3: Forming the Matrix
We can represent the vectors as rows of a matrix \( A \):

\[ A = \begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{pmatrix} \]

### Step 4: Determinant of the Matrix
The determinant of a matrix formed by orthonormal vectors (which have unit length and are mutually orthogonal) is equal to the product of their lengths (which is 1 for each vector) and the volume of the parallelepiped formed by these vectors.

Since the vectors are orthonormal, the determinant can be computed as:

\[ D = \text{Volume} = \sqrt{1^2 \cdot 1^2 \cdot 1^2} = 1 \]

### Conclusion
Thus, the value of the determinant is:

\[ \boxed{1} \]