Formula : Plane = r . (n) = d

Where r = any random point

n = normal vector of plane

d = distance of plane from origin

If two planes are parallel , then their normal vectors are same.

Therefore ,

Parallel Plane r . (2i – 3j + 5k) + 5 = 0

Normal vector = (2i - 3j + 5k)

∴ Normal vector of required plane = (2i - 3j + 5k)

Equation of required plane r . (2i - 3j + 5k) = d

In cartesian form 2x – 3y + 5z = d

Plane passes through point (3,4, - 1) therefore it will satisfy it.

2(3) – 3(4) + 5( - 1) = d

6 – 12 - 5 = d

d = - 11

Equation of required plane r . (2i - 3j + 5k) = - 11

r . (2i - 3j + 5k) + 11 = 0