Coplanar lines and plane containing them

Method: put the second line as the intersection of two planes, take the plane of the pencil through it which contains a point of the first line, and check that the first line's direction lies in that plane.

Line 1: (x+5)/3 = (y+4)/1 = (z−7)/(−2): point A(−5, −4, 7), direction (3, 1, −2). Line 2: 3x + 2y + z − 2 = 0 = x − 3y + 2z − 13.

Plane through line 2 and the point A

Any plane through line 2:

(3x + 2y + z − 2) + k(x − 3y + 2z − 13) = 0.

Put A(−5, −4, 7):

(−15 − 8 + 7 − 2) + k(−5 + 12 + 14 − 13) = 0
−18 + 8k = 0   →   k = 9/4

Plane: 4(3x + 2y + z − 2) + 9(x − 3y + 2z − 13) = 0, i.e.

21x − 19y + 22z − 125 = 0

The first line lies in this plane

The plane contains A. Its normal is (21, −19, 22) and the direction of line 1 is (3, 1, −2):

21(3) − 19(1) + 22(−2) = 63 − 19 − 44 = 0.

So the direction is perpendicular to the normal; line 1 is parallel to the plane and passes through a point of it, hence it lies in the plane. Both lines are in the same plane, so they are coplanar.

Answer: the lines are coplanar, and the plane containing them is 21x − 19y + 22z − 125 = 0.