`L_1=hati+hatj+2hatk+lamda(hati-hatj+2hatk)lamdainR
L_2=hatj-hatk+mu(3hati+hatj+phatk)
L_3=s(lhati+mhatj+nhatk)
L_1` is perpendicular to `L_2`
`L_3` is perpendicular to `L_1` & `L_2`
then (l, m, n) can be

Options

A. (-1,7,4)
B. (4, -1, 7)
C. (7, 4, -1)
D. (7, -1, 4)

Answer

To solve the problem, we need to determine the values of \( l, m, n \) such that the lines \( L_1, L_2, \) and \( L_3 \) satisfy the given conditions of perpendicularity.

### Step-by-step Solution:

1. **Identify the Direction Vectors:**
- For line \( L_1 \):
\[ L_1 = \hat{i} + \hat{j} + 2\hat{k} + \lambda(\hat{i} - \hat{j} + 2\hat{k}) \]
The direction vector \( \mathbf{b_1} \) can be expressed as:
\[ \mathbf{b_1} = \hat{i} - \hat{j} + 2\hat{k} \]
- For line \( L_2 \):
\[ L_2 = \hat{j} - \hat{k} + \mu(3\hat{i} + \hat{j} + p\hat{k}) \]
The direction vector \( \mathbf{b_2} \) can be expressed as:
\[ \mathbf{b_2} = 3\hat{i} + \hat{j} + p\hat{k} \]
- For line \( L_3 \):
\[ L_3 = s(l\hat{i} + m\hat{j} + n\hat{k}) \]
The direction vector \( \mathbf{b_3} \) can be expressed as:
\[ \mathbf{b_3} = l\hat{i} + m\hat{j} + n\hat{k} \]

2. **Condition for Perpendicularity \( L_1 \perp L_2 \):**
- The dot product of \( \mathbf{b_1} \) and \( \mathbf{b_2} \) must equal zero:
\[ \mathbf{b_1} \cdot \mathbf{b_2} = (1)(3) + (-1)(1) + (2)(p) = 0 \]
Simplifying this gives:
\[ 3 - 1 + 2p = 0 \implies 2 + 2p = 0 \implies p = -1 \]

3. **Condition for Perpendicularity \( L_3 \perp L_1 \) and \( L_2 \):**
- Since \( L_1 \) and \( L_2 \) are perpendicular, we can use their cross product to find a vector that is perpendicular to both.
- Calculate the cross product \( \mathbf{b_1} \times \mathbf{b_2} \):
\[ \mathbf{b_1} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}, \quad \mathbf{b_2} = \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix} \]
Using the determinant to find the cross product:
\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 3 & 1 & -1 \end{vmatrix} \]
Expanding this determinant:
\[ = \hat{i}((-1)(-1) - (2)(1)) - \hat{j}((1)(-1) - (2)(3)) + \hat{k}((1)(1) - (-1)(3)) \]
\[ = \hat{i}(1 - 2) - \hat{j}(-1 - 6) + \hat{k}(1 + 3) \]
\[ = -\hat{i} + 7\hat{j} + 4\hat{k} \]

4. **Set \( \mathbf{b_3} \) parallel to \( \mathbf{b_1} \times \mathbf{b_2} \):**
- Since \( L_3 \) is parallel to the cross product, we have:
\[ l\hat{i} + m\hat{j} + n\hat{k} = k(-\hat{i} + 7\hat{j} + 4\hat{k}) \quad \text{for some scalar } k \]
This gives us the equations:
\[ l = -k, \quad m = 7k, \quad n = 4k \]
- Choosing \( k = 1 \) for simplicity, we find:
\[ l = -1, \quad m = 7, \quad n = 4 \]

### Final Result:
Thus, the values of \( (l, m, n) \) can be:
\[ (l, m, n) = (-1, 7, 4) \]