If the line, `(x-3)/(1) =(y+2)/(-1) = (z+ lamda)/(-2)` lies in the plane, `2x -4y + 3z =2,` then the shortest distance between this line and the line,` (x-1)/(12) = y/9 =z/4` is :

Options

A. 0
B. 3
C. 1
D. 2

Answer

To find the shortest distance between the two lines given in the problem, we will follow these steps:

### Step 1: Identify the lines and their parameters
The first line is given by the equation:
\[ \frac{x-3}{1} = \frac{y+2}{-1} = \frac{z+\lambda}{-2} \]
This means the line passes through the point \( (3, -2, -\lambda) \) and has direction ratios \( (1, -1, -2) \).

The second line is given by:
\[ \frac{x-1}{12} = \frac{y}{9} = \frac{z}{4} \]
This line passes through the point \( (1, 0, 0) \) and has direction ratios \( (12, 9, 4) \).

### Step 2: Determine the value of \( \lambda \)
Since the first line lies in the plane defined by the equation:
\[ 2x - 4y + 3z = 2 \]
We can substitute the point \( (3, -2, -\lambda) \) into the plane equation to find \( \lambda \):
\[ 2(3) - 4(-2) + 3(-\lambda) = 2 \]
Calculating this gives:
\[ 6 + 8 - 3\lambda = 2 \]
\[ 14 - 3\lambda = 2 \]
\[ -3\lambda = 2 - 14 \]
\[ -3\lambda = -12 \]
\[ \lambda = 4 \]

### Step 3: Write the coordinates of the points and direction ratios
Now that we have \( \lambda = 4 \), the first line can be expressed as passing through the point \( (3, -2, -4) \) with direction ratios \( (1, -1, -2) \).

The second line passes through the point \( (1, 0, 0) \) with direction ratios \( (12, 9, 4) \).

### Step 4: Use the formula for the shortest distance between two skew lines
The formula for the shortest distance \( d \) between two lines is given by:
\[ d = \frac{|(P_1 - P_2) \cdot (D_1 \times D_2)|}{|D_1 \times D_2|} \]
Where:
- \( P_1 = (3, -2, -4) \) (point on the first line)
- \( P_2 = (1, 0, 0) \) (point on the second line)
- \( D_1 = (1, -1, -2) \) (direction ratios of the first line)
- \( D_2 = (12, 9, 4) \) (direction ratios of the second line)

### Step 5: Calculate \( P_1 - P_2 \)
\[ P_1 - P_2 = (3 - 1, -2 - 0, -4 - 0) = (2, -2, -4) \]

### Step 6: Calculate the cross product \( D_1 \times D_2 \)
\[ D_1 \times D_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & -2 \\ 12 & 9 & 4 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}((-1)(4) - (-2)(9)) - \hat{j}(1(4) - (-2)(12)) + \hat{k}(1(9) - (-1)(12)) \]
\[ = \hat{i}(-4 + 18) - \hat{j}(4 + 24) + \hat{k}(9 + 12) \]
\[ = \hat{i}(14) - \hat{j}(28) + \hat{k}(21) \]
So, \( D_1 \times D_2 = (14, -28, 21) \).

### Step 7: Calculate the magnitude of \( D_1 \times D_2 \)
\[ |D_1 \times D_2| = \sqrt{14^2 + (-28)^2 + 21^2} = \sqrt{196 + 784 + 441} = \sqrt{1421} \]

### Step 8: Calculate the dot product \( (P_1 - P_2) \cdot (D_1 \times D_2) \)
\[ (P_1 - P_2) \cdot (D_1 \times D_2) = (2, -2, -4) \cdot (14, -28, 21) \]
Calculating this gives:
\[ = 2(14) + (-2)(-28) + (-4)(21) = 28 + 56 - 84 = 0 \]

### Step 9: Calculate the shortest distance
Now substituting back into the distance formula:
\[ d = \frac{|0|}{\sqrt{1421}} = 0 \]

### Final Answer
The shortest distance between the two lines is \( 0 \).