Prove lines coplanar

Lines.

• L₁: (x − 1)/2 = (y − 2)/3 = (z − 3)/4, through A(1, 2, 3) with direction (2, 3, 4).
• L₂: 4x − 3y + 1 = 0 and 5x − 3z + 2 = 0.

Symmetric form of L₂

From 4x − 3y + 1 = 0: y = (4x + 1)/3 = 4x/3 + 1/3. From 5x − 3z + 2 = 0: z = (5x + 2)/3.

Put x = 3m: x = 3m, y = 4m + 1/3, z = 5m + 2/3. So L₂ passes through C(0, 1/3, 2/3) with direction (3, 4, 5). In symmetric form:

x/3 = (y − 1/3)/4 = (z − 2/3)/5.

Coplanarity condition

Two lines through A and C with directions d₁ and d₂ are coplanar if [AC, d₁, d₂] = 0.

AC = (0 − 1, 1/3 − 2, 2/3 − 3) = (−1, −5/3, −7/3) ∝ (−3, −5, −7) (multiplying by 3).

| −3  −5  −7 |
|  2   3   4 | = −3(3·5 − 4·4) + 5(2·5 − 4·3) − 7(2·4 − 3·3)
|  3   4   5 |
             = −3(−1) + 5(−2) − 7(−1)
             = 3 − 10 + 7
             = 0

The scalar triple product is zero, so the lines are coplanar. ∎

(They also intersect, at (−1, −1, −1): this is the point of L₁ with parameter −1, and it satisfies both planes of L₂.)