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 $\left|d\right|$ is equal to …

Equation of the plane containing the lines

$$\begin{gathered} \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5} \\ and \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} is a(x-2)+b(y-3)+c(z-4)=0 \\ where, 3a+4b+5c=0 \\ 2a+3b+4c=0 \\ and a(1-2)+b(2-3)+c(2-3)=0 \\ i.e, a+b+c=0 \end{gathered}$$

From Eqs. (ii) and (iii), $\frac{a}{1}=\frac{b}{-2}=\frac{c}{11},$ which satisfy Eq. (iv).
$\therefore$ Planes must be parallel, so $A=1$ and then 
$$\begin{gathered} \frac{\left|d\right|}{\sqrt{6}}=\sqrt{6} \\ \Rightarrow \left|d\right|=6 \end{gathered}$$