If `l_(1), m_(1), n_(1), l_(2), m_(2), n_(2) and l_(3), m_(3), n_(3)` are direction cosines of three mutuallyy perpendicular lines then, the value of `|(l_(1),m_(1),n_(1)),(l_(2),m_(2),n_(2)),(l_(3),m_(3),n_(3))|` is

Options

A. `l_(3) m_(3) n_(3)`
B. `+- 1`
C. `l_(1) m_(1) n_(1)`
D. `l_(2) m_(2) n_(2)`

Explanation

To solve the problem, we need to find the value of the determinant formed by the direction cosines of three mutually perpendicular lines. Let's denote the direction cosines as follows:

- For the first line: \( l_1, m_1, n_1 \)
- For the second line: \( l_2, m_2, n_2 \)
- For the third line: \( l_3, m_3, n_3 \)

### Step 1: Understand the properties of direction cosines
For three mutually perpendicular lines, the following conditions hold:

1. \( l_1^2 + m_1^2 + n_1^2 = 1 \)
2. \( l_2^2 + m_2^2 + n_2^2 = 1 \)
3. \( l_3^2 + m_3^2 + n_3^2 = 1 \)
4. \( l_1l_2 + m_1m_2 + n_1n_2 = 0 \)
5. \( l_1l_3 + m_1m_3 + n_1n_3 = 0 \)
6. \( l_2l_3 + m_2m_3 + n_2n_3 = 0 \)

### Step 2: Set up the determinant
We need to evaluate the determinant:

\[ \Delta = \begin{vmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \end{vmatrix} \]

### Step 3: Calculate the square of the determinant
To find \( \Delta^2 \), we can multiply the determinant by itself:

\[ \Delta^2 = \begin{vmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \end{vmatrix} \cdot \begin{vmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \end{vmatrix} \]

### Step 4: Expand the determinant
Using the properties of determinants, we can expand this to find:

\[ \Delta^2 = \begin{vmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ 1 & 1 & 1 \end{vmatrix} \cdot \begin{vmatrix} l_1 & m_1 & n_1 \\ l_3 & m_3 & n_3 \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 5: Substitute the known values
From the properties of direction cosines, we know that:

- The first determinant gives us \( 1 \) (since \( l_1^2 + m_1^2 + n_1^2 = 1 \)).
- The second determinant gives us \( 0 \) (since the lines are mutually perpendicular).

Thus, we can conclude:

\[ \Delta^2 = 1 \cdot 0 = 0 \]

### Step 6: Solve for \( \Delta \)
Since \( \Delta^2 = 1 \), we have:

\[ \Delta = \pm 1 \]

### Final Answer
Therefore, the value of the determinant is:

\[ \Delta = \pm 1 \]