The user wants to find the value of $|d|$ given the distance between two planes.

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

The two lines are: Line 1 ($L_1$): ${{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over 4}$ A point on this line is $P_1(1, 2, 3)$ and its direction vector is $\vec{d_1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.

Line 2 ($L_2$): ${{x - 2} \over 3} = {{y - 3} \over 4} = {{z - 4} \over 5}$ A point on this line is $P_2(2, 3, 4)$ and its direction vector is $\vec{d_2} = 3\hat{i} + 4\hat{j} + 5\hat{k}$.

The plane containing these two lines will have a normal vector $\vec{n}$ that is perpendicular to both direction vectors. We can find $\vec{n}$ by taking the cross product of $\vec{d_1}$ and $\vec{d_2}$.

$\vec{n} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}$ $\vec{n} = \hat{i}(3 \times 5 - 4 \times 4) - \hat{j}(2 \times 5 - 3 \times 4) + \hat{k}(2 \times 4 - 3 \times 3)$ $\vec{n} = \hat{i}(15 - 16) - \hat{j}(10 - 12) + \hat{k}(8 - 9)$ $\vec{n} = -\hat{i} + 2\hat{j} - \hat{k}$

The equation of the plane can be found using the normal vector $\vec{n} = (-1, 2, -1)$ and a point on the plane, for example, $P_1(1, 2, 3)$. The equation of the plane is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$. $-1(x - 1) + 2(y - 2) - 1(z - 3) = 0$ $-x + 1 + 2y - 4 - z + 3 = 0$ $-x + 2y - z = 0$ Multiplying by -1, we get the equation of the plane containing the lines as: $x - 2y + z = 0$

Step 2: Compare the two planes.

The given plane is $P_{given}: Ax - 2y + z = d$. The plane we found is $P_{lines}: x - 2y + z = 0$.

The problem states that the distance between these two planes is $\sqrt{6}$. For a non-zero distance to exist between two planes, they must be parallel. Two planes are parallel if their normal vectors are proportional.

The normal vector of $P_{given}$ is $\vec{n_1} = (A, -2, 1)$. The normal vector of $P_{lines}$ is $\vec{n_2} = (1, -2, 1)$.

For the planes to be parallel, $\frac{A}{1} = \frac{-2}{-2} = \frac{1}{1}$, which implies $A=1$. So, the given plane is $x - 2y + z = d$.

Step 3: Calculate the distance and find $|d|$.

The distance $D$ between two parallel planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$ is given by the formula: $D = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}$ In our case, the two planes are $x - 2y + z = d$ and $x - 2y + z = 0$. So, $a=1, b=-2, c=1, d_1=d, d_2=0$.

$D = \frac{|d - 0|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{|d|}{\sqrt{1 + 4 + 1}} = \frac{|d|}{\sqrt{6}}$ We are given that the distance is $\sqrt{6}$. $\frac{|d|}{\sqrt{6}} = \sqrt{6}$ $|d| = \sqrt{6} \times \sqrt{6}$ $|d| = 6$

The value of $|d|$ is 6.