If the distance between the plane αx – 2y + z = k and the plane containing the lines \(\rm \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\ \)and \(\rm \frac{x-2}{3}=\frac{y-3}{4}=\frac {z-4}{5}\) is √6, then |k| is

1
36
2
12
3
6
4
2√3

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