If the distance between the plane $Ax - 2y + z = d$ and the plane containing the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}$ is $\sqrt{6}$, then find $|d|$.

$P_1:\ Ax - 2y + z = d$ $L_1:\ \vec r = (1,2,3) + \lambda(2,3,4), \qquad L_2:\ \vec r = (2,3,4) + \mu(3,4,5)$ $P_2 \text{ contains both } L_1 \text{ and } L_2$ $\text{dist}(P_1, P_2) = \sqrt{6} \quad \Longrightarrow \quad \text{find } |d|$ $L_1 \Rightarrow A(1,2,3), \quad \vec{d_1} = (2,3,4)$ $L_2 \Rightarrow B(2,3,4), \quad \vec{d_2} = (3,4,5)$ $\text{Plan: } 1.\ \vec{d_1}\times\vec{d_2} = \text{normal of } P_2$ $2.\ P_2 \text{ through } A;\ \ P_1 \parallel P_2 \Rightarrow \text{coefficient } A$ $3.\ \text{distance between parallel planes} \Rightarrow |d|$ $\vec{d_1}\times\vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}$ $= \hat{i}(15-16) - \hat{j}(10-12) + \hat{k}(8-9)$ $\vec{d_1}\times\vec{d_2} = (-1,\ 2,\ -1)$ $\vec{AB} = B - A = (1,1,1)$ $\vec{AB}\cdot(\vec{d_1}\times\vec{d_2}) = -1 + 2 - 1 = 0$ $\Rightarrow L_1, L_2 \text{ are coplanar}$ $\text{normal} = (-1,2,-1) \propto (1,-2,1)$ $P_2:\ x - 2y + z = 0 \quad (\text{through } A)$ $P_1:\ Ax - 2y + z = d \Rightarrow \text{normal } (A,-2,1)$ $(A,-2,1) \parallel (1,-2,1) \Rightarrow A = 1$ $P_1:\ x - 2y + z = d$ $d(P_1,P_2) = \dfrac{|d - 0|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \dfrac{|d|}{\sqrt{6}}$ $\dfrac{|d|}{\sqrt{6}} = \sqrt{6}$ $\boxed{|d| = 6}$ $|d| = 6$