A point on the first line can be found where t=0: (x, y, z) = (1, 2, 0).
A point on the second line can be found where s=0: (2, 6, -2).
A line perpendicular to both the given lines will have a direction vector that is the cross product of the direction vectors of the given lines:
... (1, 6, 2) × (2, 14, 5) = (2, -1, 2)
A unit vector in this direction will be this vector divided by its magnitude:
... (2, -1, 2)/√(2²+(-1)²+2²) = (2/3, -1/3, 2/3)
The projection of any vector between the two given lines onto this unit vector will have a length equal to the distance between the lines. That is, we can find the distance between the skew lines by forming the dot product of the vector between the two points first found and this unit direction vector.
... (1-2, 2-6, 0-(-2))•(2/3, -1/3, 2/3)
... = -2/3 +4/3 +4/3
... = 2
The distance between the given lines is 2 units.
