Since the plane is parallel to 2x – 3y + 5z + 7 = 0, it must be of the form:

2x – 3y + 5z + θ = 0

According to question,

The plane passes through (3, 4, –1)

⟹ 2(3) – 3(4) +5(–1) + θ = 0

⟹ θ = 11

So, the equation of the plane is as follows:

2x – 3y + 5z + 11 = 0

Distance of the plane 2x – 3y + 5z + 7 = 0 from (3, 4, –1):

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

is given by:

Putting the necessary values,

⟹

⟹

∴ the distance of the plane 2x – 3y + 5z + 7 = 0 from (3, 4, –1) is

units