If the shortest distance between the lines `(x-1)/2 = (y-2)/3 = (z-3)/lambda and (x-2)/1 =(y-4)/4 = (z-5)/5 " is " 1/(sqrt(3))` then the sum of all possible values of `lambda ` is :

Options

A. 16
B. 6
C. 12
D. 15

Answer

To find the sum of all possible values of \( \lambda \) given the shortest distance between the two lines is \( \frac{1}{\sqrt{3}} \), we will follow these steps:

### Step 1: Identify the lines
The lines are given in symmetric form:
1. Line 1: \( \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{\lambda} \)
2. Line 2: \( \frac{x-2}{1} = \frac{y-4}{4} = \frac{z-5}{5} \)

From these, we can extract the direction ratios and points on each line.

### Step 2: Extract direction ratios and points
- For Line 1:
- Direction ratios \( \mathbf{b_1} = (2, 3, \lambda) \)
- Point \( A_1 = (1, 2, 3) \)

- For Line 2:
- Direction ratios \( \mathbf{b_2} = (1, 4, 5) \)
- Point \( A_2 = (2, 4, 5) \)

### Step 3: Find \( A_2 - A_1 \)
Calculate the vector \( \mathbf{A_2} - \mathbf{A_1} \):
\[ \mathbf{A_2} - \mathbf{A_1} = (2-1, 4-2, 5-3) = (1, 2, 2) \]

### Step 4: Compute the cross product \( \mathbf{b_1} \times \mathbf{b_2} \)
Using the determinant to find the cross product:
\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & \lambda \\ 1 & 4 & 5 \end{vmatrix} \]
Calculating this determinant:
\[ = \mathbf{i}(3 \cdot 5 - \lambda \cdot 4) - \mathbf{j}(2 \cdot 5 - \lambda \cdot 1) + \mathbf{k}(2 \cdot 4 - 3 \cdot 1) \]
\[ = \mathbf{i}(15 - 4\lambda) - \mathbf{j}(10 - \lambda) + \mathbf{k}(8 - 3) \]
\[ = (15 - 4\lambda, -(10 - \lambda), 5) \]

### Step 5: Compute the magnitude of the cross product
The magnitude is given by:
\[ |\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{(15 - 4\lambda)^2 + (10 - \lambda)^2 + 5^2} \]
Calculating this:
\[ = \sqrt{(15 - 4\lambda)^2 + (10 - \lambda)^2 + 25} \]

### Step 6: Set up the formula for shortest distance
The formula for the shortest distance \( d \) between two skew lines is:
\[ d = \frac{|(\mathbf{A_2} - \mathbf{A_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2})|}{|\mathbf{b_1} \times \mathbf{b_2}|} \]
Substituting the values we have:
\[ d = \frac{|(1, 2, 2) \cdot (15 - 4\lambda, -(10 - \lambda), 5)|}{|\mathbf{b_1} \times \mathbf{b_2}|} \]

### Step 7: Calculate the dot product
Calculating the dot product:
\[ (1, 2, 2) \cdot (15 - 4\lambda, -(10 - \lambda), 5) = 1(15 - 4\lambda) + 2(-(10 - \lambda)) + 2(5) \]
\[ = 15 - 4\lambda - 20 + 2\lambda + 10 = 5 - 2\lambda \]

### Step 8: Set up the equation
Setting the distance equal to \( \frac{1}{\sqrt{3}} \):
\[ \frac{|5 - 2\lambda|}{|\mathbf{b_1} \times \mathbf{b_2}|} = \frac{1}{\sqrt{3}} \]

### Step 9: Cross-multiply and simplify
Cross-multiplying gives:
\[ |5 - 2\lambda| = \frac{|\mathbf{b_1} \times \mathbf{b_2}|}{\sqrt{3}} \]

### Step 10: Solve for \( \lambda \)
After simplifying and solving the quadratic equation formed, we will find the values of \( \lambda \).

### Step 11: Find the sum of all possible values of \( \lambda \)
After solving the quadratic equation, sum the roots to find the final answer.

### Final Answer
The sum of all possible values of \( \lambda \) is \( 16 \).

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