Consider the following 3lines in space
`L_1:r=3hat(i)-hat(j)+hat(k)+lambda(2hat(i)+4hat(j)-hat(k))`
`L_2: r=hat(i)+hat(j)-3hat(k)+mu(4hat(i)+2hat(j)+4hat(k))`
`L_3:=3hat(i)+2hat(j)-2hat(k)+t(2hat(i)+hat(j)+2hat(k))`
Then, which one of the following part(s) is/ are in the same plane?

Options

A. Only `L_1L_2`
B. Only `L_2L_3`
C. Only `L_1L_3`
D. `L_1L_2 and L_2L_3`

Answer

To determine which lines among \( L_1 \), \( L_2 \), and \( L_3 \) lie in the same plane, we can use the concept of the scalar triple product. The lines will be coplanar if the determinant formed by their direction vectors and the vector connecting points on the lines is zero.

### Step 1: Identify the lines and their components
The lines are given as:
- \( L_1: \mathbf{r} = (3\hat{i} - \hat{j} + \hat{k}) + \lambda(2\hat{i} + 4\hat{j} - \hat{k}) \)
- \( L_2: \mathbf{r} = (\hat{i} + \hat{j} - 3\hat{k}) + \mu(4\hat{i} + 2\hat{j} + 4\hat{k}) \)
- \( L_3: \mathbf{r} = (3\hat{i} + 2\hat{j} - 2\hat{k}) + t(2\hat{i} + \hat{j} + 2\hat{k}) \)

From these equations, we can extract:
- For \( L_1 \):
- Point \( A = (3, -1, 1) \)
- Direction vector \( \mathbf{b} = (2, 4, -1) \)

- For \( L_2 \):
- Point \( C = (1, 1, -3) \)
- Direction vector \( \mathbf{d} = (4, 2, 4) \)

- For \( L_3 \):
- Point \( B = (3, 2, -2) \)
- Direction vector \( \mathbf{e} = (2, 1, 2) \)

### Step 2: Check coplanarity of \( L_1 \) and \( L_2 \)
To check if \( L_1 \) and \( L_2 \) are in the same plane, we need to calculate the determinant of the matrix formed by the direction vectors and the vector connecting points \( C \) and \( A \).

1. Calculate \( \mathbf{c} - \mathbf{a} \):
\[ \mathbf{c} - \mathbf{a} = (1 - 3, 1 - (-1), -3 - 1) = (-2, 2, -4) \]

2. Form the determinant:
\[ \text{Det} = \begin{vmatrix} 2 & 4 & -1 \\ 4 & 2 & 4 \\ -2 & 2 & -4 \end{vmatrix} \]

3. Calculate the determinant:
\[ = 2 \begin{vmatrix} 2 & 4 \\ 2 & -4 \end{vmatrix} - 4 \begin{vmatrix} 4 & 4 \\ -2 & -4 \end{vmatrix} - 1 \begin{vmatrix} 4 & 2 \\ -2 & 2 \end{vmatrix} \]

\[ = 2(2 \cdot -4 - 4 \cdot 2) - 4(4 \cdot -4 - 4 \cdot -2) - (4 \cdot 2 - 2 \cdot -2) \]
\[ = 2(-8 - 8) - 4(-16 + 8) - (8 + 4) \]
\[ = 2(-16) - 4(-8) - 12 \]
\[ = -32 + 32 - 12 = -12 \neq 0 \]

Thus, \( L_1 \) and \( L_2 \) are **not** in the same plane.

### Step 3: Check coplanarity of \( L_1 \) and \( L_3 \)
1. Calculate \( \mathbf{b} - \mathbf{a} \):
\[ \mathbf{b} - \mathbf{a} = (3 - 3, 2 - (-1), -2 - 1) = (0, 3, -3) \]

2. Form the determinant:
\[ \text{Det} = \begin{vmatrix} 2 & 4 & -1 \\ 2 & 1 & 2 \\ 0 & 3 & -3 \end{vmatrix} \]

3. Calculate the determinant:
\[ = 2 \begin{vmatrix} 1 & 2 \\ 3 & -3 \end{vmatrix} - 4 \begin{vmatrix} 2 & 2 \\ 0 & -3 \end{vmatrix} - 1 \begin{vmatrix} 2 & 1 \\ 0 & 3 \end{vmatrix} \]

\[ = 2(-3 - 6) - 4(-6) - (6) \]
\[ = 2(-9) + 24 - 6 \]
\[ = -18 + 24 - 6 = 0 \]

Thus, \( L_1 \) and \( L_3 \) are in the same plane.

### Step 4: Check coplanarity of \( L_2 \) and \( L_3 \)
1. Calculate \( \mathbf{c} - \mathbf{b} \):
\[ \mathbf{c} - \mathbf{b} = (1 - 3, 1 - 2, -3 - (-2)) = (-2, -1, -1) \]

2. Form the determinant:
\[ \text{Det} = \begin{vmatrix} 4 & 2 & 4 \\ 2 & 1 & 2 \\ -2 & -1 & -1 \end{vmatrix} \]

3. Calculate the determinant:
\[ = 4 \begin{vmatrix} 1 & 2 \\ -1 & -1 \end{vmatrix} - 2 \begin{vmatrix} 2 & 4 \\ -2 & -1 \end{vmatrix} + 4 \begin{vmatrix} 2 & 1 \\ -2 & -1 \end{vmatrix} \]

\[ = 4(-1 + 2) - 2(-2 - 8) + 4(-2 + 2) \]
\[ = 4(1) + 2(10) + 0 \]
\[ = 4 + 20 = 24 \neq 0 \]

Thus, \( L_2 \) and \( L_3 \) are **not** in the same plane.

### Conclusion
The only lines that are in the same plane are \( L_1 \) and \( L_3 \).