If the distance between the plane ax 2y + z = d and the plane containing the lines `(x-1)/2=(y-2)/3=(z-3)/4` and `(x-2)/3=(y-3)/4=(z-4)/5` is `sqrt6` , then value of |d| is

Explanation

To solve the problem step by step, we need to find the value of |d| given the distance between two planes. Let's break it down:

### Step 1: Identify the equations of the planes
The first plane is given as:
\[ ax + 2y + z = d \]

The second plane is determined by the lines:
1. \(\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}\)
2. \(\frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}\)

### Step 2: Find the direction vectors of the lines
From the equations of the lines, we can extract the direction vectors:
- For the first line, the direction vector \( \mathbf{n_1} = \langle 2, 3, 4 \rangle \)
- For the second line, the direction vector \( \mathbf{n_2} = \langle 3, 4, 5 \rangle \)

### Step 3: Find the normal vector of the plane containing the lines
To find the normal vector of the plane containing these lines, we take the cross product of the two direction vectors:
\[ \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} \]

Calculating the determinant:
\[ = \mathbf{i}(3 \cdot 5 - 4 \cdot 4) - \mathbf{j}(2 \cdot 5 - 4 \cdot 3) + \mathbf{k}(2 \cdot 4 - 3 \cdot 3) \]
\[ = \mathbf{i}(15 - 16) - \mathbf{j}(10 - 12) + \mathbf{k}(8 - 9) \]
\[ = -\mathbf{i} + 2\mathbf{j} - \mathbf{k} \]
Thus, the normal vector of the plane containing the lines is:
\[ \mathbf{n} = \langle -1, 2, -1 \rangle \]

### Step 4: Write the equation of the plane
Using the point (1, 2, 3) from the first line, the equation of the plane can be written as:
\[ -1(x - 1) + 2(y - 2) - 1(z - 3) = 0 \]
Simplifying this gives:
\[ -x + 1 + 2y - 4 - z + 3 = 0 \]
\[ -x + 2y - z = 0 \]
or
\[ x - 2y + z = 0 \]

### Step 5: Find the distance between the two planes
The distance \(d\) between two parallel planes \(Ax + By + Cz = D_1\) and \(Ax + By + Cz = D_2\) is given by:
\[ \text{Distance} = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}} \]

Here, for the planes \(ax + 2y + z = d\) and \(x - 2y + z = 0\):
- For the first plane, \(A = a\), \(B = 2\), \(C = 1\), and \(D_1 = d\).
- For the second plane, \(D_2 = 0\).

The distance is given as \(\sqrt{6}\):
\[ \sqrt{6} = \frac{|d - 0|}{\sqrt{a^2 + 2^2 + 1^2}} = \frac{|d|}{\sqrt{a^2 + 4 + 1}} = \frac{|d|}{\sqrt{a^2 + 5}} \]

### Step 6: Set up the equation
From the above, we have:
\[ |d| = \sqrt{6} \cdot \sqrt{a^2 + 5} \]
Squaring both sides gives:
\[ d^2 = 6(a^2 + 5) \]

### Step 7: Find |d|
To find |d|, we need to express it in terms of \(a\):
\[ |d| = \sqrt{6a^2 + 30} \]

### Conclusion
The value of |d| will depend on the value of \(a\). However, since we are not given \(a\), we can conclude that the expression for |d| is:
\[ |d| = \sqrt{6a^2 + 30} \]