If the distance between the plane $x-2y+z=d$ 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 $\sqrt{6},$ then find |d|.

$L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$

$L_2: \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$

$\pi: x-2y+z=d$

$\text{Given: distance between the two planes } = \sqrt{6}$

$\text{To find: } |d|$

$A_1(1,2,3),\ \vec{a}=(2,3,4)$

$A_2(2,3,4),\ \vec{b}=(3,4,5)$

$\text{Plan: normal of the plane through both lines } = \vec{a}\times\vec{b}$

$\text{Write that plane } \pi_1 \text{, then compare with } x-2y+z=d$

$\text{Use the distance } \sqrt{6} \Rightarrow |d|$

n⃗ = a⃗×b⃗ = î ĵ k̂ 2 3 4 3 4 5

$= \hat{i}(15-16) - \hat{j}(10-12) + \hat{k}(8-9)$

$\vec{n} = (-1,\ 2,\ -1)$

$A_2 - A_1 = (2-1,\ 3-2,\ 4-3) = (1,\ 1,\ 1)$

$\vec{A_1A_2}\cdot\vec{n} = 1(-1) + 1(2) + 1(-1) = 0$

$\Rightarrow L_1, L_2 \text{ are coplanar}$

$\pi_1: (x-1) - 2(y-2) + (z-3) = 0$

$\pi_1: x - 2y + z = 0$

$\pi: x-2y+z=d \Rightarrow \vec{n}_\pi = (1,-2,1)$

$\vec{n}_{\pi_1} = \vec{n}_\pi \Rightarrow \pi \parallel \pi_1$

$\text{distance} = \frac{|d_1-d_2|}{\sqrt{A^2+B^2+C^2}}$

$= \frac{|d-0|}{\sqrt{1^2+(-2)^2+1^2}} = \frac{|d|}{\sqrt{6}}$

$\frac{|d|}{\sqrt{6}} = \sqrt{6}$

$|d| = 6$

$|d| = 6 \Rightarrow d = \pm 6$

$\text{Answer: } |d| = 6$