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 2x – 3y + 7z + 13 = 0

Normal vector = (2i - 3j + 7k)

∴ Normal vector of required plane = (2i - 3j + 7k)

Equation of required plane r . (2i - 3j + 7k) = d

In cartesian form 2x - 3y + 7z = d

Plane passes through point (0,0,0) therefore it will satisfy it.

2(0) - (0) + 3(0) = d

d = 0

Equation of required plane 2x - 3y + 7z = 0