If `l_1^2+m_1^2+n_1^2=1` etc. and `l_1l_2+m_1m_2+n_1n_2=0` etc. then
`Delta=|(l_1,m_1,n_1),(l_2,m_2,n_2),(l_3,m_3,n_3)|=`

Options

A. 1
B. 2
C. 3
D. `pm1`

Answer

To solve the problem, we need to find the value of the determinant \( \Delta = |(l_1, m_1, n_1), (l_2, m_2, n_2), (l_3, m_3, n_3)| \) given the conditions \( l_1^2 + m_1^2 + n_1^2 = 1 \), \( l_2^2 + m_2^2 + n_2^2 = 1 \), \( l_3^2 + m_3^2 + n_3^2 = 1 \), and the orthogonality conditions \( l_1 l_2 + m_1 m_2 + n_1 n_2 = 0 \), \( l_1 l_3 + m_1 m_3 + n_1 n_3 = 0 \), \( l_2 l_3 + m_2 m_3 + n_2 n_3 = 0 \).

### Step-by-step Solution:

1. **Understanding the Determinant**:
The determinant \( \Delta \) represents the volume of the parallelepiped formed by the vectors \( (l_1, m_1, n_1) \), \( (l_2, m_2, n_2) \), and \( (l_3, m_3, n_3) \).

2. **Using Properties of Determinants**:
We know that if the vectors are orthogonal and each vector has a magnitude of 1, then the volume of the parallelepiped they form is equal to the product of their magnitudes. Since all vectors are unit vectors and orthogonal, we can conclude that the determinant will be either 1 or -1.

3. **Setting Up the Determinant**:
The determinant can be expressed as:
\[ \Delta = \begin{vmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \end{vmatrix} \]

4. **Calculating the Determinant**:
We can use the property of determinants that states if we take the transpose of a matrix, the determinant remains unchanged. Thus, we can consider the determinant of the transpose:
\[ \Delta^2 = \begin{vmatrix} l_1 & l_2 & l_3 \\ m_1 & m_2 & m_3 \\ n_1 & n_2 & n_3 \end{vmatrix} \cdot \begin{vmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \end{vmatrix} \]

5. **Expanding the Determinant**:
When we expand the determinant, we will find that the diagonal elements will contribute 1 (since each vector is a unit vector), and the off-diagonal elements will be 0 due to the orthogonality conditions.

6. **Final Calculation**:
Thus, we find:
\[ \Delta^2 = 1 \]
Therefore, taking the square root gives:
\[ \Delta = \pm 1 \]

### Conclusion:
The value of the determinant \( \Delta \) is either \( 1 \) or \( -1 \).