Distance Between Plane and Plane Containing Two Lines

To find the value of $|d|$, we first determine the equation of the plane containing the two given lines:
$L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$
$L_2: \frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}$

Line $L_1$ passes through the point $P_1(1, 2, 3)$ with direction vector $\vec{v}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
Line $L_2$ passes through the point $P_2(2, 3, 4)$ with direction vector $\vec{v}_2 = 3\hat{i} + 4\hat{j} + 5\hat{k}$.

The normal vector $\vec{n}$ to the plane containing both lines is given by the cross product of their direction vectors:

$$\vec{n} = \vec{v}_1 \times \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}$$

Evaluating the determinant:
$\vec{n} = \hat{i}(15 - 16) - \hat{j}(10 - 12) + \hat{k}(8 - 9) = -\hat{i} + 2\hat{j} - \hat{k}$

Multiplying by $-1$, we can take the normal vector as:
$\vec{n} = \hat{i} - 2\hat{j} + \hat{k}$

We verify that the lines are coplanar by checking if the vector connecting points $P_1$ and $P_2$, $\vec{P_1P_2} = \langle 2-1, 3-2, 4-3 \rangle = \langle 1, 1, 1 \rangle$, is perpendicular to $\vec{n}$:
$\vec{P_1P_2} \cdot \vec{n} = 1(1) + 1(-2) + 1(1) = 0$
Since the dot product is zero, the two lines are indeed coplanar.

The equation of the plane containing these lines is:
$1(x - 1) - 2(y - 2) + 1(z - 3) = 0$
$x - 2y + z - 1 + 4 - 3 = 0$
$x - 2y + z = 0$

We are given another plane:
$Ax - 2y + z = d$

For the distance between the two planes to be a non-zero finite constant, the planes must be parallel. Comparing the direction ratios of the normals:
$A = 1$

Thus, the two parallel planes are:
$x - 2y + z = 0 \quad \text{and} \quad x - 2y + z = d$

The distance between these two parallel planes is given by:
$\text{Distance} = \frac{|d - 0|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{|d|}{\sqrt{6}}$

Given that the distance is $\sqrt{6}$:
$\frac{|d|}{\sqrt{6}} = \sqrt{6}$
$|d| = 6$