We have a plane

$ Ax - 2y + z = d $

and another plane that contains the two lines

$ \text{Line 1:}\quad \frac{x - 1}{2} \;=\; \frac{y - 2}{3} \;=\; \frac{z - 3}{4}, $

$ \text{Line 2:}\quad \frac{x - 2}{3} \;=\; \frac{y - 3}{4} \;=\; \frac{z - 4}{5}. $

We know the distance between these two planes is $\sqrt{6}$, and we want to find $\lvert d\rvert.$

1. Find the equation of the plane containing the two given lines

Step 1a: Parametric forms of the lines

Line 1: Let the parameter be $t$. Then

$ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \;=\; t \;\;\Longrightarrow\;\; \begin{cases} x = 1 + 2t,\\ y = 2 + 3t,\\ z = 3 + 4t. \end{cases} $

A direction vector for Line 1 is $\mathbf{v}_1 = (2,\,3,\,4)$.

A point on Line 1 is $\mathbf{P}_1 = (1,2,3)$.

Line 2: Let the parameter be $s$. Then

$ \frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5} \;=\; s \;\;\Longrightarrow\;\; \begin{cases} x = 2 + 3s,\\ y = 3 + 4s,\\ z = 4 + 5s. \end{cases} $

A direction vector for Line 2 is $\mathbf{v}_2 = (3,\,4,\,5)$.

A point on Line 2 is $\mathbf{P}_2 = (2,3,4)$.

Step 1b: Normal to the plane containing these lines

A plane that contains both lines must contain their direction vectors $\mathbf{v}_1$ and $\mathbf{v}_2$. Therefore, a normal to this plane is given by the cross product $\mathbf{v}_1 \times \mathbf{v}_2$.

$ \mathbf{v}_1 = (2,\,3,\,4), \quad \mathbf{v}_2 = (3,\,4,\,5). $

Compute the cross product:

$ \mathbf{v}_1 \times \mathbf{v}_2 = \det\!\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 2 & 3 & 4\\ 3 & 4 & 5 \end{pmatrix} = \bigl(3\cdot 5 - 4\cdot 4,\; 4\cdot 3 - 2\cdot 5,\; 2\cdot 4 - 3\cdot 3\bigr) = (15 - 16,\; 12 - 10,\; 8 - 9) = (-1,\; 2,\; -1). $

Hence a normal vector to the plane is $\mathbf{n} = (-1,\,2,\,-1)$. Equivalently, we can multiply by $-1$ (which does not change the plane) to get $\mathbf{n} = (1, -2, 1)$.

Thus the plane containing the two lines has the form

$ 1\cdot x \;-\; 2\cdot y \;+\; 1\cdot z \;=\; K, $

i.e.

$ x \;-\; 2y \;+\; z \;=\; K. $

Step 1c: Find the constant $K$

To find $K$, just plug in any point on either line. For instance, the point $\mathbf{P}_1 = (1,2,3)$ on Line 1:

$ 1(1)\;-\;2(2)\;+\;1(3) \;=\; 1 -4 +3 = 0. $

So $K=0$.

Check also with $\mathbf{P}_2 = (2,3,4)$ from Line 2:

$ 2 - 2\cdot 3 + 4 = 2 -6 +4 = 0. $

That also gives $0$. So indeed the plane containing both lines is

$ \boxed{x - 2y + z = 0}. $

2. Determine $A$ so that the planes can be parallel

We are given the plane

$ Ax - 2y + z = d $

and have found that the plane containing the lines is

$ x - 2y + z = 0. $

For these two planes to have a finite, nonzero distance between them, they must be parallel. Two planes are parallel precisely when their normal vectors are scalar multiples of each other.

The normal to $Ax - 2y + z = d$ is $\,(A,\,-2,\,1)$.

The normal to $x - 2y + z = 0$ is $\,(1,\,-2,\,1)$.

Set

$ (A,\,-2,\,1) \;=\; \lambda\,\bigl(1,\,-2,\,1\bigr). $

Matching components:

$A = \lambda\cdot 1 = \lambda$.

$-2 = \lambda\cdot (-2)$ $\implies$ $\lambda = 1$.

$1 = \lambda\cdot (1)$ $\implies$ $\lambda = 1$.

Hence $\lambda=1$ and $A=1$.

Therefore, the given plane must be

$ \boxed{x - 2y + z = d}. $

3. Use the formula for the distance between two parallel planes

Now we have two parallel planes:

$x - 2y + z = 0$,

$x - 2y + z = d$.

The normal vector to both is $\mathbf{n} = (1, -2, 1)$. Its magnitude is

$ \|\mathbf{n}\| \;=\; \sqrt{\,1^2 + (-2)^2 + 1^2\,} \;=\; \sqrt{\,1 + 4 + 1\,} \;=\; \sqrt{6}. $

The distance $D$ between two parallel planes

$ \alpha_1: \quad \mathbf{n}\cdot\mathbf{x} = k_1, \quad \alpha_2: \quad \mathbf{n}\cdot\mathbf{x} = k_2 $

is given by

$ D \;=\; \frac{\lvert k_1 - k_2\rvert}{\|\mathbf{n}\|}. $

In our case:

For the plane $x - 2y + z = 0$, we have $k_1 = 0$.

For the plane $x - 2y + z = d$, we have $k_2 = d$.

The distance is given to be $\sqrt{6}$.

Thus

$ \sqrt{6} \;=\; \frac{\lvert 0 - d\rvert}{\sqrt{6}} \;=\; \frac{\lvert d\rvert}{\sqrt{6}} \;\;\Longrightarrow\;\; \lvert d\rvert = 6. $

4. Conclusion

$ \boxed{\lvert d\rvert = 6}. $