The vector equations of two lines `L_1 and L_2` are respectively, `L_1:r=2i+9j+13k+lambda(i+2j+3k) and L_2: r=-3i+7j+pk +mu(-i+2j-3k)` Then, the lines `L_1 and L_2` are

Options

A. skew lines all `p in R`
B. intersecting for all `p in R` and the point of intersection is `(-1, 3, 4)`
C. intersecting lines for `p=-2`
D. intersecting for all real `p in R`

Answer

To determine the relationship between the two lines \( L_1 \) and \( L_2 \), we first need to analyze their vector equations.

### Step 1: Write down the vector equations of the lines
The vector equations of the lines are given as:
- \( L_1: \mathbf{r} = 2\mathbf{i} + 9\mathbf{j} + 13\mathbf{k} + \lambda(\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) \)
- \( L_2: \mathbf{r} = -3\mathbf{i} + 7\mathbf{j} + p\mathbf{k} + \mu(-\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}) \)

### Step 2: Rewrite the equations in parametric form
For \( L_1 \):
- \( x_1 = 2 + \lambda \)
- \( y_1 = 9 + 2\lambda \)
- \( z_1 = 13 + 3\lambda \)

For \( L_2 \):
- \( x_2 = -3 - \mu \)
- \( y_2 = 7 + 2\mu \)
- \( z_2 = p - 3\mu \)

### Step 3: Set the equations equal to find intersection points
For the lines to intersect, the corresponding components must be equal:
1. \( 2 + \lambda = -3 - \mu \) (Equation 1)
2. \( 9 + 2\lambda = 7 + 2\mu \) (Equation 2)
3. \( 13 + 3\lambda = p - 3\mu \) (Equation 3)

### Step 4: Solve the first two equations
From Equation 1:
\[ \lambda + \mu = -5 \quad \text{(Rearranging gives us Equation A)} \]

From Equation 2:
\[ 2\lambda - 2\mu = -2 \quad \text{(Rearranging gives us Equation B)} \]
Dividing by 2:
\[ \lambda - \mu = -1 \quad \text{(Equation B)} \]

### Step 5: Solve the system of equations (A and B)
Now we have:
1. \( \lambda + \mu = -5 \) (A)
2. \( \lambda - \mu = -1 \) (B)

Adding these two equations:
\[ 2\lambda = -6 \implies \lambda = -3 \]

Substituting \( \lambda = -3 \) into Equation A:
\[ -3 + \mu = -5 \implies \mu = -2 \]

### Step 6: Substitute \( \lambda \) and \( \mu \) into Equation 3
Now substitute \( \lambda = -3 \) and \( \mu = -2 \) into Equation 3:
\[ 13 + 3(-3) = p - 3(-2) \]
\[ 13 - 9 = p + 6 \]
\[ 4 = p + 6 \implies p = -2 \]

### Conclusion
The lines \( L_1 \) and \( L_2 \) intersect at the point where \( p = -2 \).

### Final Answer
The lines \( L_1 \) and \( L_2 \) intersect at \( p = -2 \).

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