If the shortest distance between the lines `frac{x - lambda} {-2}` = `frac{y - 2} {1}`=`frac{z - 1} {1}` and `frac{x - sqrt 3}{1}` = `frac{y-1}{-2}`=`frac{z-2}{1}` is 1, then the sum of all possible values of `lambda` is :

Options

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

Answer

To solve the problem, we need to find the sum of all possible values of \( \lambda \) such that the shortest distance between the two given lines is equal to 1.

### Step 1: Write the equations of 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 expressed 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 expressed in vector form 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 points and direction vectors

From the above, we identify:
- Point on line 1: \( \mathbf{A_1} = \begin{pmatrix} \lambda \\ 2 \\ 1 \end{pmatrix} \)
- Direction vector of line 1: \( \mathbf{B_1} = \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix} \)
- Point on line 2: \( \mathbf{A_2} = \begin{pmatrix} \sqrt{3} \\ 1 \\ 2 \end{pmatrix} \)
- Direction vector of line 2: \( \mathbf{B_2} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \)

### Step 3: Calculate \( \mathbf{A_2} - \mathbf{A_1} \)

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

### Step 4: Calculate \( \mathbf{B_1} \times \mathbf{B_2} \)

Next, we compute the cross product \( \mathbf{B_1} \times \mathbf{B_2} \):
\[ \mathbf{B_1} \times \mathbf{B_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 1 & 1 \\ 1 & -2 & 1 \end{vmatrix} \]
Calculating the determinant:
\[ = \mathbf{i} \begin{vmatrix} 1 & 1 \\ -2 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} -2 & 1 \\ 1 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} -2 & 1 \\ 1 & -2 \end{vmatrix} \]
\[ = \mathbf{i} (1 + 2) - \mathbf{j} (-2 - 1) + \mathbf{k} (4 - 1) \]
\[ = 3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k} = 3\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \]

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

The magnitude of \( \mathbf{B_1} \times \mathbf{B_2} \) is:
\[ |\mathbf{B_1} \times \mathbf{B_2}| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt{3} \]

### Step 6: Use the formula for the shortest distance between two skew lines

The formula for the shortest distance \( d \) between two skew lines is given by:
\[ d = \frac{|(\mathbf{A_2} - \mathbf{A_1}) \cdot (\mathbf{B_1} \times \mathbf{B_2})|}{|\mathbf{B_1} \times \mathbf{B_2}|} \]
Substituting the values we have:
\[ d = \frac{\left| \begin{pmatrix} \sqrt{3} - \lambda \\ -1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 3 \\ 3 \end{pmatrix} \right|}{3\sqrt{3}} \]
Calculating the dot product:
\[ = \left| 3(\sqrt{3} - \lambda) - 3 + 3 \right| = \left| 3(\sqrt{3} - \lambda) \right| \]
Thus,
\[ d = \frac{3|\sqrt{3} - \lambda|}{3\sqrt{3}} = \frac{|\sqrt{3} - \lambda|}{\sqrt{3}} \]

### Step 7: Set the distance equal to 1 and solve for \( \lambda \)

Setting the distance equal to 1:
\[ \frac{|\sqrt{3} - \lambda|}{\sqrt{3}} = 1 \]
This implies:
\[ |\sqrt{3} - \lambda| = \sqrt{3} \]
This gives us two cases:
1. \( \sqrt{3} - \lambda = \sqrt{3} \) → \( \lambda = 0 \)
2. \( \sqrt{3} - \lambda = -\sqrt{3} \) → \( \lambda = 2\sqrt{3} \)

### Step 8: Find the sum of all possible values of \( \lambda \)

The possible values of \( \lambda \) are \( 0 \) and \( 2\sqrt{3} \). Therefore, the sum is:
\[ 0 + 2\sqrt{3} = 2\sqrt{3} \]

### Final Answer
The sum of all possible values of \( \lambda \) is \( 2\sqrt{3} \).