Coplanar lines and plane containing them

Two lines are coplanar if the vector joining a point on each is perpendicular to the cross product of their directions, i.e. the scalar triple product is zero.

Line 1: (x+1)/(−3) = (y−3)/2 = (z+2)/1, point A(−1, 3, −2), direction a = (−3, 2, 1). Line 2: x/1 = (y−7)/(−3) = (z+7)/2, point B(0, 7, −7), direction b = (1, −3, 2).

Proof of coplanarity

AB = B − A = (1, 4, −5).

| 1   4  −5 |
|−3   2   1 |  = 1(4+3) − 4(−6−1) + (−5)(9−2)
| 1  −3   2 |  = 7 + 28 − 35 = 0

The determinant is zero, so the lines are coplanar. (They are not parallel since a and b are not proportional, so they intersect.)

Equation of the plane

The normal is perpendicular to both directions:

a × b = (2·2 − 1·(−3), 1·1 − (−3)·2, (−3)(−3) − 2·1)
      = (7, 7, 7)  ∥  (1, 1, 1)

The plane through A with normal (1, 1, 1):

(x + 1) + (y − 3) + (z + 2) = 0, i.e. x + y + z = 0.

Check: B(0, 7, −7) gives 0 + 7 − 7 = 0, so the second line's point is also in the plane.

Answer: the lines are coplanar and lie in the plane x + y + z = 0.