Coplanar lines, intersection point and plane

How I read the question. As printed, the first line is (x − 5)/4 = (y − 7)/4 = (z − 3)/(−5). With this line the two lines are not coplanar (the determinant below is 144, not 0). The standard textbook version has (z + 3)/(−5), and the coplanarity then works exactly, so I solve that intended version:

• L₁: (x − 5)/4 = (y − 7)/4 = (z + 3)/(−5), through A(5, 7, −3), direction d₁ = (4, 4, −5).
• L₂: (x − 8)/7 = (y − 4)/1 = (z − 5)/3, through B(8, 4, 5), direction d₂ = (7, 1, 3).

Coplanarity

AB = (3, −3, 8).

| 3  −3   8 |
| 4   4  −5 | = 3(12 + 5) + 3(12 + 35) + 8(4 − 28)
| 7   1   3 |
            = 51 + 141 − 192
            = 0

The determinant is zero, so the lines are coplanar. (They are not parallel, since (4, 4, −5) is not proportional to (7, 1, 3).)

Point of intersection

L₁: (5 + 4t, 7 + 4t, −3 − 5t); L₂: (8 + 7s, 4 + s, 5 + 3s).

5 + 4t = 8 + 7s    →  4t − 7s = 3
7 + 4t = 4 + s     →  4t − s = −3

Subtract: −6s = 6, so s = −1; then 4t = −3 − 1 = −4, t = −1.

Third coordinate: −3 − 5(−1) = 2 and 5 + 3(−1) = 2 ✓.

Point: (5 − 4, 7 − 4, 2) = (1, 3, 2).

Equation of the plane

Normal = d₁ × d₂ = (4, 4, −5) × (7, 1, 3) = (12 + 5, −35 − 12, 4 − 28) = (17, −47, −24).

Through (1, 3, 2): 17(x − 1) − 47(y − 3) − 24(z − 2) = 0.

17x − 47y − 24z − 17 + 141 + 48 = 0
17x − 47y − 24z + 172 = 0

Check with A(5, 7, −3): 85 − 329 + 72 + 172 = 0 ✓; with B(8, 4, 5): 136 − 188 − 120 + 172 = 0 ✓.

Answer: the lines meet at (1, 3, 2); the plane is 17x − 47y − 24z + 172 = 0.