The vectors A = ⟨5, -4, 4⟩, B = ⟨-5, 0, -3⟩, and C = ⟨-4, 2, -5⟩ point to the points A, B, and C in the plane respectively.
Take any two of these vectors and compute their difference. This new vector will point in a direction parallel to the plane. (If it's unclear why, consider the triangle law of vector addition.) We'll need two such vectors. Here, I'll take
B - A = ⟨-5, 0, -3⟩ - ⟨5, -4, 4⟩ = ⟨-10, 4, -7⟩
and
C - A = ⟨-4, 2, -5⟩ - ⟨5, -4, 4⟩ = ⟨-9, 6, -9⟩
but you would get the same end result if you picked A - B and C - B, or A - C and B - C.
The cross product of any two vectors is perpendicular to both of those vectors. Both B - A and C - A point parallel to the same plane, so their cross product will be perpendicular to that plane.
(B - A) × (C - A) = ⟨-10, 4, -7⟩×⟨-9, 6, -9⟩ = ⟨6, -27, -24⟩
This answer is not unique, since any multiple of this vector will be perpendicular or normal to the given plane. For example, since each component is a multiple of 3, you can simplify the normal vector to ⟨2, -9, -8⟩.
In case you're unsure of how the cross product was computed: recall the definition of ×,
⟨1, 0, 0⟩×⟨0, 1, 0⟩ = ⟨0, 0, 1⟩
⟨0, 1, 0⟩×⟨0, 0, 1⟩ = ⟨1, 0, 0⟩
⟨0, 0, 1⟩×⟨1, 0, 0⟩ = ⟨0, 1, 0⟩
For any vector v, we have v × v = 0, the zero vector. Also, for any two vectors u and v, we have u × v = -(v × u).
Expand both vectors in terms of the unit vectors,
⟨-10, 4, -7⟩ = -10 ⟨1, 0, 0⟩ + 4 ⟨0, 1, 0⟩ - 7 ⟨0, 0, 1⟩
⟨-9, 6, -9⟩ = -9 ⟨1, 0, 0⟩ + 6 ⟨0, 1, 0⟩ - 9 ⟨0, 0, 1⟩
The cross product is distributive, so that
⟨-10, 4, -7⟩×⟨-9, 6, -9⟩
= (-10 ⟨1, 0, 0⟩ + 4 ⟨0, 1, 0⟩ - 7 ⟨0, 0, 1⟩) × (-9 ⟨1, 0, 0⟩ + 6 ⟨0, 1, 0⟩ - 9 ⟨0, 0, 1⟩)
= 90 ⟨1, 0, 0⟩×⟨1, 0, 0⟩ - 60 ⟨1, 0, 0⟩×⟨0, 1, 0⟩ + 90 ⟨1, 0, 0⟩×⟨0, 0, 1⟩
… - 36 ⟨0, 1, 0⟩×⟨1, 0, 0⟩ + 24 ⟨0, 1, 0⟩×⟨0, 1, 0⟩ - 36 ⟨0, 1, 0⟩×⟨0, 0, 1⟩
… + 63 ⟨0, 0, 1⟩×⟨1, 0, 0⟩ - 42 ⟨0, 0, 1⟩×⟨0, 1, 0⟩ + 63 ⟨0, 0, 1⟩×⟨0, 0, 1⟩
= -60 ⟨0, 0, 1⟩ - 90 ⟨0, 1, 0⟩ + 36 ⟨0, 0, 1⟩ - 36 ⟨1, 0, 0⟩ + 63 ⟨0, 1, 0⟩ + 42 ⟨1, 0, 0⟩
= -24 ⟨0, 0, 1⟩ - 27 ⟨0, 1, 0⟩ + 6 ⟨1, 0, 0⟩
= ⟨6, -27, -24⟩
