If the lines `(x-2)/(1)=(y-3)/(1)=(z-4)/(lamda)` and `(x-1)/(lamda)=(y-4)/(2)=(z-5)/(1)` intersect then

Options

A. `lamda=-1`
B. `lamda=2`
C. `lamda=-3`
D. `lamda=0`

Explanation

To solve the problem of whether the two lines intersect, we will follow a systematic approach.

### Step-by-Step Solution:

1. **Parameterization of the Lines:**
- For the first line, we can set:
\[ \frac{x-2}{1} = \frac{y-3}{1} = \frac{z-4}{\lambda} = t_1 \]
From this, we can express \(x\), \(y\), and \(z\) in terms of \(t_1\):
\[ x = t_1 + 2, \quad y = t_1 + 3, \quad z = \lambda t_1 + 4 \]

- For the second line, we set:
\[ \frac{x-1}{\lambda} = \frac{y-4}{2} = \frac{z-5}{1} = t_2 \]
From this, we express \(x\), \(y\), and \(z\) in terms of \(t_2\):
\[ x = \lambda t_2 + 1, \quad y = 2t_2 + 4, \quad z = t_2 + 5 \]

2. **Equating the Expressions for \(x\), \(y\), and \(z\):**
- Equate the expressions for \(x\):
\[ t_1 + 2 = \lambda t_2 + 1 \quad \Rightarrow \quad t_1 - \lambda t_2 = -1 \quad \text{(Equation 1)} \]

- Equate the expressions for \(y\):
\[ t_1 + 3 = 2t_2 + 4 \quad \Rightarrow \quad t_1 - 2t_2 = 1 \quad \text{(Equation 2)} \]

- Equate the expressions for \(z\):
\[ \lambda t_1 + 4 = t_2 + 5 \quad \Rightarrow \quad \lambda t_1 - t_2 = 1 \quad \text{(Equation 3)} \]

3. **Solving the Equations:**
- From Equation 2, we can express \(t_1\) in terms of \(t_2\):
\[ t_1 = 2t_2 + 1 \]

- Substitute \(t_1\) into Equation 1:
\[ (2t_2 + 1) - \lambda t_2 = -1 \]
Simplifying this gives:
\[ 2t_2 + 1 - \lambda t_2 = -1 \quad \Rightarrow \quad (2 - \lambda)t_2 = -2 \quad \Rightarrow \quad t_2 = \frac{-2}{2 - \lambda} \]

- Substitute \(t_1\) into Equation 3:
\[ \lambda(2t_2 + 1) - t_2 = 1 \]
Simplifying this gives:
\[ 2\lambda t_2 + \lambda - t_2 = 1 \quad \Rightarrow \quad (2\lambda - 1)t_2 = 1 - \lambda \]
Thus, we have:
\[ t_2 = \frac{1 - \lambda}{2\lambda - 1} \]

4. **Equating the Two Expressions for \(t_2\):**
- Set the two expressions for \(t_2\) equal to each other:
\[ \frac{-2}{2 - \lambda} = \frac{1 - \lambda}{2\lambda - 1} \]

- Cross-multiplying gives:
\[ -2(2\lambda - 1) = (1 - \lambda)(2 - \lambda) \]
Expanding both sides:
\[ -4\lambda + 2 = 2 - 3\lambda + \lambda^2 \]
Rearranging leads to:
\[ \lambda^2 + \lambda - 4 = 0 \]

5. **Finding the Roots:**
- Using the quadratic formula:
\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 16}}{2} = \frac{-1 \pm \sqrt{17}}{2} \]

### Final Result:
The values of \(\lambda\) that allow the lines to intersect are:
\[ \lambda = 0 \quad \text{or} \quad \lambda = 2 \]