If the lines `(x-4)/1=(y-2)/1=(z-lamda)/3` and `x/1=(y+2)/2=z/4` intersect each other, then `lamda` lies in the interval

Options

A. (9,11)
B. (-5,-3)
C. (13,15)
D. (11,13)

Explanation

To determine the interval in which the value of \( \lambda \) lies, given that the lines

\[ \frac{x-4}{1} = \frac{y-2}{1} = \frac{z-\lambda}{3} \]

and

\[ \frac{x}{1} = \frac{y+2}{2} = \frac{z}{4} \]

intersect, we will use the condition for coplanarity of two lines.

### Step-by-Step Solution

1. **Identify Points and Direction Ratios:**
- For the first line, we can extract the point and direction ratios:
- Point \( P_1(4, 2, \lambda) \)
- Direction ratios \( \mathbf{a} = (1, 1, 3) \)
- For the second line:
- Point \( P_2(0, -2, 0) \)
- Direction ratios \( \mathbf{b} = (1, 2, 4) \)

2. **Set Up the Determinant for Coplanarity:**
- The condition for coplanarity of the two lines is given by the determinant:
\[ \begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = 0 \]
- Substituting the values:
\[ \begin{vmatrix} 0 - 4 & -2 - 2 & 0 - \lambda \\ 1 & 1 & 3 \\ 1 & 2 & 4 \end{vmatrix} = 0 \]
- This simplifies to:
\[ \begin{vmatrix} -4 & -4 & -\lambda \\ 1 & 1 & 3 \\ 1 & 2 & 4 \end{vmatrix} = 0 \]

3. **Calculate the Determinant:**
- Expanding the determinant:
\[ -4 \begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix} + 4 \begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} - \lambda \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} \]
- Calculating each 2x2 determinant:
- \( \begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix} = (1)(4) - (3)(2) = 4 - 6 = -2 \)
- \( \begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} = (1)(4) - (3)(1) = 4 - 3 = 1 \)
- \( \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = (1)(2) - (1)(1) = 2 - 1 = 1 \)

4. **Substituting Back into the Determinant:**
- Now substituting these values back:
\[ -4(-2) + 4(1) - \lambda(1) = 0 \]
- This simplifies to:
\[ 8 + 4 - \lambda = 0 \]
- Therefore:
\[ \lambda = 12 \]

5. **Determine the Interval:**
- The question asks for the interval in which \( \lambda \) lies. Since \( \lambda = 12 \), we need to check the provided intervals.
- The correct interval is \( (3, 15) \) since \( 12 \) lies within this range.

### Final Answer:
The value of \( \lambda \) lies in the interval \( (3, 15) \).