Question

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(6, -2, 0)


(5, -2, 3)


(-8, 3, 4)


(2, 5, -1)

Answer 1

First find the plane through the first three points A(6,-2,0), B(5,-2,3), C(-8,3,4).

Vectors AB = B − A = (−1, 0, 3), AC = C − A = (−14, 5, 4).
Normal vector n = AB × AC = (15, 38, 5) (you can take the negative of the cross product computed).

Plane equation: 15x + 38y + 5z + d = 0.
Plug A(6,−2,0): 15·6 + 38·(−2) + 5·0 + d = 90 − 76 + d = 14 + d = 0 ⇒ d = −14.

Thus the plane is 15x + 38y + 5z − 14 = 0.

Test D(2,5,−1): 15·2 + 38·5 + 5·(−1) − 14 = 30 + 190 − 5 − 14 = 201 ≠ 0.

Therefore the fourth point does not lie in the plane determined by the first three points — the four points are not coplanar.