The equation of the plane passing through (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) is given by the following equation:

According to question,

(x₁, y₁, z₁) = (2, 5, –3)

(x₂, y₂, z₂) = (–2, –3, 5)

(x₃, y₃, z₃) = (5, 3, –3)

Putting these values,

⟹

⟹ (x – 2)(16) + (y – 5)(24) + (z + 3)(32) = 0

⟹ 2x + 3y + 4z –7 = 0

Distance of 2x + 3y + 4z –7 = 0 from (7, 2, 4)

We know, the distance of point (x₁,y₁,z₁) from the plane

is given by:

⟹

⟹ p = √29 units

∴ the distance between the point (7, 2, 4) and the plane determined by the points A(2, 5, –3), B(–2, –3, 5) and (5, 3, –3) is √29 units.