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 determine the value of \( \lambda \) for which the given lines intersect, we can use the condition that the shortest distance between the two lines is zero. This can be expressed using the formula involving the direction ratios of the lines.

### Step-by-Step Solution:

1. **Identify the Direction Ratios and Points:**
The first line is given by:
\[ \frac{x-2}{1} = \frac{y-3}{1} = \frac{z-4}{\lambda} \]
This can be rewritten in parametric form as:
\[ x = 2 + t, \quad y = 3 + t, \quad z = 4 + \lambda t \]
where \( t \) is a parameter.

The second line is given by:
\[ \frac{x-1}{\lambda} = \frac{y-4}{2} = \frac{z-5}{1} \]
This can be rewritten in parametric form as:
\[ x = 1 + \lambda s, \quad y = 4 + 2s, \quad z = 5 + s \]
where \( s \) is another parameter.

2. **Find Points on Each Line:**
From the first line, we can denote a point \( A(2, 3, 4) \) and the direction ratios as \( (1, 1, \lambda) \).
From the second line, we can denote a point \( B(1, 4, 5) \) and the direction ratios as \( (\lambda, 2, 1) \).

3. **Use the Condition for Intersection:**
The lines will intersect if the vector connecting points \( A \) and \( B \) is perpendicular to the direction vectors of both lines. This can be expressed as:
\[ (B - A) \cdot (d_1 \times d_2) = 0 \]
where \( d_1 = (1, 1, \lambda) \) and \( d_2 = (\lambda, 2, 1) \).

4. **Calculate \( B - A \):**
\[ B - A = (1 - 2, 4 - 3, 5 - 4) = (-1, 1, 1) \]

5. **Calculate the Cross Product \( d_1 \times d_2 \):**
\[ d_1 \times d_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & \lambda \\ \lambda & 2 & 1 \end{vmatrix} \]
Expanding this determinant:
\[ = \hat{i}(1 \cdot 1 - \lambda \cdot 2) - \hat{j}(1 \cdot 1 - \lambda \cdot \lambda) + \hat{k}(1 \cdot 2 - 1 \cdot \lambda) \]
\[ = \hat{i}(1 - 2\lambda) - \hat{j}(1 - \lambda^2) + \hat{k}(2 - \lambda) \]

6. **Set Up the Equation:**
Now, we need to compute:
\[ (B - A) \cdot (d_1 \times d_2) = 0 \]
This gives:
\[ (-1, 1, 1) \cdot (1 - 2\lambda, -(1 - \lambda^2), 2 - \lambda) = 0 \]
Expanding this:
\[ -1(1 - 2\lambda) + 1(-1 + \lambda^2) + 1(2 - \lambda) = 0 \]
Simplifying:
\[ -1 + 2\lambda - 1 + \lambda^2 + 2 - \lambda = 0 \]
\[ \lambda^2 + \lambda = 0 \]

7. **Factor the Equation:**
\[ \lambda(\lambda + 1) = 0 \]

8. **Solve for \( \lambda \):**
This gives us:
\[ \lambda = 0 \quad \text{or} \quad \lambda = -1 \]

### Final Answer:
The values of \( \lambda \) for which the lines intersect are \( \lambda = 0 \) and \( \lambda = -1 \).