`(x-lamda)/2=(y-2)/1=(z-1)/1 "and"(x-sqrt3)/1=(y-1)/(-2)=(z-2)/1` If the shortest distance between the above two lines is 1 then sum of possible values of `lamda`.

Options

A. 0
B. `2sqrt3`
C. `3sqrt3`
D. `-2sqrt3`

Answer

To solve the problem, we need to find the sum of possible values of \(\lambda\) given that the shortest distance between the two lines is equal to 1. Let's go through the solution step by step.

### Step 1: Represent the lines in vector form

The first line is given by:
\[ \frac{x - \lambda}{2} = \frac{y - 2}{1} = \frac{z - 1}{1} \]
This can be represented in vector form as:
\[ \mathbf{R_1} = \begin{pmatrix} \lambda \\ 2 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} \]
where \(t\) is a parameter.

The second line is given by:
\[ \frac{x - \sqrt{3}}{1} = \frac{y - 1}{-2} = \frac{z - 2}{1} \]
This can be represented as:
\[ \mathbf{R_2} = \begin{pmatrix} \sqrt{3} \\ 1 \\ 2 \end{pmatrix} + s \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \]
where \(s\) is another parameter.

### Step 2: Identify the direction vectors and points

From the equations, we can identify:
- For line 1:
- Point \(A_1 = \begin{pmatrix} \lambda \\ 2 \\ 1 \end{pmatrix}\)
- Direction vector \(B_1 = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\)

- For line 2:
- Point \(A_2 = \begin{pmatrix} \sqrt{3} \\ 1 \\ 2 \end{pmatrix}\)
- Direction vector \(B_2 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\)

### Step 3: Calculate \(A_2 - A_1\)

Now, we calculate the vector \(A_2 - A_1\):
\[ A_2 - A_1 = \begin{pmatrix} \sqrt{3} - \lambda \\ 1 - 2 \\ 2 - 1 \end{pmatrix} = \begin{pmatrix} \sqrt{3} - \lambda \\ -1 \\ 1 \end{pmatrix} \]

### Step 4: Calculate the cross product \(B_1 \times B_2\)

Next, we calculate the cross product of the direction vectors \(B_1\) and \(B_2\):
\[ B_1 \times B_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 1 \\ 1 & -2 & 1 \end{vmatrix} \]
Calculating this determinant:
\[ = \mathbf{i} \left( 1 \cdot 1 - 1 \cdot (-2) \right) - \mathbf{j} \left( 2 \cdot 1 - 1 \cdot 1 \right) + \mathbf{k} \left( 2 \cdot (-2) - 1 \cdot 1 \right) \]
\[ = \mathbf{i} (1 + 2) - \mathbf{j} (2 - 1) + \mathbf{k} (-4 - 1) \]
\[ = 3\mathbf{i} - 1\mathbf{j} - 5\mathbf{k} \]
Thus,
\[ B_1 \times B_2 = \begin{pmatrix} 3 \\ -1 \\ -5 \end{pmatrix} \]

### Step 5: Calculate the magnitude of \(B_1 \times B_2\)

The magnitude of the cross product is:
\[ |B_1 \times B_2| = \sqrt{3^2 + (-1)^2 + (-5)^2} = \sqrt{9 + 1 + 25} = \sqrt{35} \]

### Step 6: Use the formula for the shortest distance

The formula for the shortest distance \(D\) between two skew lines is given by:
\[ D = \frac{|(A_2 - A_1) \cdot (B_1 \times B_2)|}{|B_1 \times B_2|} \]
Setting this equal to 1:
\[ \frac{|(\sqrt{3} - \lambda, -1, 1) \cdot (3, -1, -5)|}{\sqrt{35}} = 1 \]

### Step 7: Calculate the dot product

Calculating the dot product:
\[ (\sqrt{3} - \lambda) \cdot 3 + (-1) \cdot (-1) + 1 \cdot (-5) = 3(\sqrt{3} - \lambda) + 1 - 5 \]
\[ = 3\sqrt{3} - 3\lambda - 4 \]

### Step 8: Set up the equation

Setting the absolute value equal to \(\sqrt{35}\):
\[ |3\sqrt{3} - 3\lambda - 4| = \sqrt{35} \]

This gives us two equations to solve:
1. \(3\sqrt{3} - 3\lambda - 4 = \sqrt{35}\)
2. \(3\sqrt{3} - 3\lambda - 4 = -\sqrt{35}\)

### Step 9: Solve the equations

**For the first equation:**
\[ 3\sqrt{3} - 4 - \sqrt{35} = 3\lambda \]
\[ \lambda = \frac{3\sqrt{3} - 4 - \sqrt{35}}{3} \]

**For the second equation:**
\[ 3\sqrt{3} - 4 + \sqrt{35} = 3\lambda \]
\[ \lambda = \frac{3\sqrt{3} - 4 + \sqrt{35}}{3} \]

### Step 10: Calculate the sum of possible values of \(\lambda\)

Now, we sum the two values of \(\lambda\):
\[ \text{Sum} = \frac{(3\sqrt{3} - 4 - \sqrt{35}) + (3\sqrt{3} - 4 + \sqrt{35})}{3} \]
\[ = \frac{2(3\sqrt{3} - 4)}{3} = \frac{6\sqrt{3} - 8}{3} \]

### Final Answer

Thus, the sum of possible values of \(\lambda\) is:
\[ \frac{6\sqrt{3} - 8}{3} \]