If `I_r^2+m_r^2+n_r^2=1` where r=1,2,3 and `I_1I_2+m_1m_2+n_1n_2=0` …..etc. then
`Delta=|{:(I_1,m_1,n_1),(I_2,m_2,n_2),(I_3,m_3,n_3):}|^2` is equal to

Options

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

Answer

To solve the problem, we need to evaluate the determinant \( \Delta = |(I_1, m_1, n_1), (I_2, m_2, n_2), (I_3, m_3, n_3)|^2 \) given the conditions:

1. \( I_r^2 + m_r^2 + n_r^2 = 1 \) for \( r = 1, 2, 3 \)
2. \( I_1 I_2 + m_1 m_2 + n_1 n_2 = 0 \)
3. \( I_2 I_3 + m_2 m_3 + n_2 n_3 = 0 \)
4. \( I_1 I_3 + m_1 m_3 + n_1 n_3 = 0 \)

### Step 1: Form the Matrix
We can represent the vectors as rows in a matrix \( A \):
\[ A = \begin{pmatrix} I_1 & m_1 & n_1 \\ I_2 & m_2 & n_2 \\ I_3 & m_3 & n_3 \end{pmatrix} \]

### Step 2: Calculate the Determinant
The determinant \( \Delta \) can be calculated as:
\[ \Delta = |A| = \begin{vmatrix} I_1 & m_1 & n_1 \\ I_2 & m_2 & n_2 \\ I_3 & m_3 & n_3 \end{vmatrix} \]

### Step 3: Use the Given Conditions
From the conditions provided:
- The first condition \( I_r^2 + m_r^2 + n_r^2 = 1 \) indicates that each vector is a unit vector.
- The second, third, and fourth conditions indicate that the vectors are mutually orthogonal.

### Step 4: Properties of Determinants
Since the vectors are orthogonal and of unit length, the matrix \( A \) is an orthogonal matrix. The determinant of an orthogonal matrix is either \( 1 \) or \( -1 \). Therefore:
\[ |A|^2 = 1^2 = 1 \]

### Step 5: Final Result
Thus, we have:
\[ \Delta = |A|^2 = 1 \]

### Conclusion
The value of \( \Delta \) is \( 1 \).