If the distance between the plane Ax 2y + z = d and the plane containing the lines 2 1x = 3 2y = 4 3z and 3 2x = 4 3y = 5 4z is 6 , then |d| is

Options

A. `3`
B. 4
C. 6
D. 1

Explanation

To solve the problem, we need to find the value of |d| given the distance between two planes. Here’s the step-by-step solution:

### Step 1: Identify the equations of the planes
We have the first plane given by the equation:
\[ Ax - 2y + z = d \]
We need to find the equation of the plane that contains the two lines given by:
1. \( \frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3} \)
2. \( \frac{x - 3}{2} = \frac{y - 4}{3} = \frac{z - 5}{4} \)

### Step 2: Find the direction ratios and a point on each line
From the first line, the direction ratios are \( (1, 2, 3) \) and a point on the line is \( (2, 3, 4) \).
From the second line, the direction ratios are \( (2, 3, 4) \) and a point on the line is \( (3, 4, 5) \).

### Step 3: Find the normal vector of the plane containing the lines
To find the equation of the plane containing both lines, we can take the cross product of the direction ratios of the two lines:
- Direction vector of the first line: \( \vec{d_1} = (1, 2, 3) \)
- Direction vector of the second line: \( \vec{d_2} = (2, 3, 4) \)

The cross product \( \vec{n} = \vec{d_1} \times \vec{d_2} \) gives us the normal vector of the plane.

### Step 4: Calculate the cross product
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{vmatrix} = \hat{i}(2 \cdot 4 - 3 \cdot 3) - \hat{j}(1 \cdot 4 - 3 \cdot 2) + \hat{k}(1 \cdot 3 - 2 \cdot 2) \]
\[ = \hat{i}(8 - 9) - \hat{j}(4 - 6) + \hat{k}(3 - 4) \]
\[ = \hat{i}(-1) + \hat{j}(2) + \hat{k}(-1) = (-1, 2, -1) \]

### Step 5: Write the equation of the plane
Using the point \( (2, 3, 4) \) and the normal vector \( (-1, 2, -1) \), the equation of the plane can be written as:
\[ -1(x - 2) + 2(y - 3) - 1(z - 4) = 0 \]
Simplifying this, we get:
\[ -x + 2y - z + 2 - 6 + 4 = 0 \implies -x + 2y - z = 0 \implies x - 2y + z = 0 \]

### Step 6: Find the distance between the two planes
Now we have the two planes:
1. \( Ax - 2y + z = d \)
2. \( x - 2y + z = 0 \)

The distance \( D \) between the two planes is given by the formula:
\[ D = \frac{|d - 0|}{\sqrt{A^2 + B^2 + C^2}} \]
Where \( A = 1, B = -2, C = 1 \).

### Step 7: Substitute values into the distance formula
Given that the distance \( D = 6 \):
\[ 6 = \frac{|d|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{|d|}{\sqrt{1 + 4 + 1}} = \frac{|d|}{\sqrt{6}} \]
Thus, we can solve for \( |d| \):
\[ |d| = 6 \cdot \sqrt{6} \]

### Step 8: Conclusion
Therefore, the value of \( |d| \) is \( 6\sqrt{6} \).