3D Geometry · Distance between parallel lines

Let the foot of perpendicular from the point

(\lambda, 2, 3)

on the line

\dfrac{x-4}{1} = \dfrac{y-9}{2} = \dfrac{z-5}{1}

be the point

(1, \mu, 2)

. Then the distance between the lines

\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z+4}{6}

and

\dfrac{x-\lambda}{2} = \dfrac{y-\mu}{3} = \dfrac{z+5}{6}

is equal to

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