Distance between a point and a Plane in 3 D

You are given a points (x1, y1, z1) and a plane a * x + b * y + c * z + d = 0. The task is to find the perpendicular(shortest) distance between that point and the given Plane.
 

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Examples : 
 

Input: x1 = 4, y1 = -4, z1 = 3, a = 2, b = -2, c = 5, d = 8 
Output: Perpendicular distance is 6.78902858227
Input: x1 = 2, y1 = 8, z1 = 5, a = 1, b = -2, c = -2, d = -1 
Output: Perpendicular distance is 8.33333333333 

Approach: The perpendicular distance (i.e shortest distance) from a given point to a Plane is the perpendicular distance from that point to the given plane. Let the co-ordinate of the given point be (x1, y1, z1) 
and equation of the plane be given by the equation a * x + b * y + c * z + d = 0, where a, b and c are real constants.
The formula for distance between a point and Plane in 3-D is given by:
 

Distance = (| a*x1 + b*y1 + c*z1 + d |) / (sqrt( a*a + b*b + c*c))

Below is the implementation of the above formulae: 

Perpendicular distance is 6.78902858227

Time complexity: O(log(a²+b²+c²)) as inbuilt sqrt function is being used
Auxiliary space: O(1)