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 3x – 5y + 4z = 11

Normal vector = (3i - 5j + 4k)

∴ Normal vector of required plane = (3i - 5j + 4k)

Equation of required plane r . (3i - 5j + 4k) = d

In cartesian form 3x - 5y + 4z = d

Plane passes through point ( - 1,0,7) therefore it will satisfy it.

3( - 1) - 5(0) + 4(7) = d

d = - 3 + 28 = 25

Equation of required plane 3x - 5y + 4z = 25