Let `L_(1):vec r=(hat i-hat j+2 hat k)+lambda(hat i-hat j+2 hat k),lambda in R`
`L_(2):vec r=(hat j-hat k)+mu(3 hat i+hat j+p hat k),mu in R` ,and `L_(3):vec r=delta(l hat i+m hat j+n hat k),delta in R` be three lines such that `L_(1)` is perpendicular to `L_(2)` and `L_(3)` is perpendicular to both `L_(1)` and `L_(2)` .Then,the point which lies on `L_(3)` is

Options

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

Answer

To solve the problem, we need to find the point that lies on the line \( L_3 \) given the conditions about the lines \( L_1 \), \( L_2 \), and \( L_3 \).

### Step-by-step Solution:

1. **Identify Direction Vectors**:
- For line \( L_1 \):
\[ \vec{r} = \hat{i} - \hat{j} + 2\hat{k} + \lambda(\hat{i} - \hat{j} + 2\hat{k}) \]
The direction vector \( \vec{d_1} \) is \( \hat{i} - \hat{j} + 2\hat{k} \).
- For line \( L_2 \):
\[ \vec{r} = \hat{j} - \hat{k} + \mu(3\hat{i} + \hat{j} + p\hat{k}) \]
The direction vector \( \vec{d_2} \) is \( 3\hat{i} + \hat{j} + p\hat{k} \).

2. **Use the Perpendicular Condition**:
- Since \( L_1 \) is perpendicular to \( L_2 \), we have:
\[ \vec{d_1} \cdot \vec{d_2} = 0 \]
This gives:
\[ (\hat{i} - \hat{j} + 2\hat{k}) \cdot (3\hat{i} + \hat{j} + p\hat{k}) = 0 \]
Expanding this, we get:
\[ 3 - 1 + 2p = 0 \implies 2 + 2p = 0 \implies p = -1 \]

3. **Substitute \( p \) back into \( L_2 \)**:
- Now, the direction vector for \( L_2 \) becomes:
\[ \vec{d_2} = 3\hat{i} + \hat{j} - \hat{k} \]

4. **Find the Cross Product**:
- To find the direction vector \( \vec{d_3} \) of line \( L_3 \), we compute the cross product \( \vec{d_1} \times \vec{d_2} \):
\[ \vec{d_1} = (1, -1, 2), \quad \vec{d_2} = (3, 1, -1) \]
The cross product is calculated as:
\[ \vec{d_3} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 3 & 1 & -1 \end{vmatrix} \]
This results in:
\[ \hat{i}((-1)(-1) - (2)(1)) - \hat{j}((1)(-1) - (2)(3)) + \hat{k}((1)(1) - (-1)(3)) \]
Simplifying this gives:
\[ \hat{i}(1 - 2) - \hat{j}(-1 - 6) + \hat{k}(1 + 3) = -\hat{i} + 7\hat{j} + 4\hat{k} \]

5. **Equation of Line \( L_3 \)**:
- The line \( L_3 \) can be expressed as:
\[ \vec{r} = \delta(-\hat{i} + 7\hat{j} + 4\hat{k}) \]
where \( \delta \in \mathbb{R} \).

6. **Finding a Point on \( L_3 \)**:
- For \( \delta = 1 \):
\[ \vec{r} = -\hat{i} + 7\hat{j} + 4\hat{k} \]
- Thus, the point that lies on \( L_3 \) is:
\[ (-1, 7, 4) \]

### Final Answer:
The point which lies on \( L_3 \) is \( (-1, 7, 4) \).