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p = p1 + t(A)

q = q1 + s(B)

these lines will be parallel if the unit direction vectors A and B are the same.

Let us define p as

p = p1 + t(p2 - p1) where

p1 = (1, 2, 0) and p2 - p1 = (2, -1, 3) taken from x = 1 + 2t, y = 2 - t, z = 3t

Let A = (p2-p1)/norm(p2-p1) be the normalized direction vector of p2-p1 where

norm(p2-p1) = sqrt(4 + 1 + 9) = sqrt(14)

Then A = (2/sqrt(14), -1/sqrt(14), 3/sqrt(14)) which is the normalized direction vector for line p so that

p = p1 + t(A)

If we set B = A, then the parametric line q passing through the point q1 = (3, -2, 1) is given by

q = q1 + s(B)

Because B = A, line q is parallel to the line p given above.

Note that when the direction vector is normalized, the parameters "s" and "t" for lines q and p represent the Euclidian distance from the points q1 and p1 respectively.

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