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

Equation of the plane containing the lines

$\frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}-3}{4}=\frac{\mathrm{z}-4}{5} \mathrm{and} \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{4}$

is $\mathrm{a}(\mathrm{x}-2)+\mathrm{b}(\mathrm{y}-3)+\mathrm{c}(\mathrm{z}-4)=0$

where, $3\mathrm{a}+4\mathrm{b}+5\mathrm{c}=0$  $2\mathrm{a}+3\mathrm{b}+4\mathrm{c}=0$

$\mathrm{and}\mathrm{a}(1-2)+\mathrm{b}(2-3)+\mathrm{c}(2-3)=0$

i.e., a + b + c = 0

From Equs.(ii) and (iii) $\frac{\mathrm{a}}{1}=\frac{\mathrm{b}}{-2}=\frac{\mathrm{c}}{1},$ which satisfy
Eq (iv) Plane through lines is $\mathrm{x}-2\mathrm{y}+\mathrm{z}=0$
Given plane is $\mathrm{Ax}-2\mathrm{y}+\mathrm{z}=\mathrm{d}\mathrm{is}\sqrt{6}$
$\therefore$ Planes must be parallel , so A = 1 and then $\frac{\left|\mathrm{d}\right|}{\sqrt{6}}=\sqrt{6}\Rightarrow \left|\mathrm{d}\right|=6$