If the distance between the plane \(\mathrm{A} x-2 y+\mathrm{z}=\mathrm{d}\) 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 \(\sqrt{6}\) units, then \(|d|\) is

Equation of the plane containing the given lines is
\(\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-2 & z-3 \\
2 & 3 & 4 \\
3 & 4 & 5
\end{array}\right|=0 \\
& \Rightarrow(x-1)(15-16)-(y-2)(10-12) \\
& +(z-3)(8-9)=0
\end{aligned}\)
\(\begin{aligned}
& \Rightarrow(x-1)(-1)-(y-2)(-2)+(z-3)(-1)=0 \\
& \Rightarrow-x+1+2 y-4-z+3=0 \\
& \Rightarrow-x+2 y-z=0 \\
& \Rightarrow x-2 y+z=0...(i)
\end{aligned}\)
Given equation of plane is
\(A x-2 y+z=d...(ii)\)
The planes given by equations (i) and (ii) are parallel.
\(\therefore \quad A=1\)
Distance between the planes ( D ) is
\(\begin{aligned} & \quad D=\left|\frac{d}{\sqrt{1^2+(-2)^2+1^2}}\right|=\left|\frac{d}{\sqrt{6}}\right| \\ & \therefore \quad\left|\frac{d}{\sqrt{6}}\right|=\sqrt{6} \\ & \quad \Rightarrow|d|=6\end{aligned}\)