Distance of a plane I have posted this question in Stack Overflow programming forum. Someone there feels it might be more suited to Mathematics. 
I have to warn you I am rusty in math and was terrible in Algebra. 
Using Accord Framework plane class. The help on this page specifies a constructor that creates a plane a certain distance from the origin. I am getting numbers that do not make sense. If I create a plane using the vector 5,5,5 at a distance of 1 then I would expect the plane to be 1 away from the origin:
        Plane p = new Plane(new Vector3(5,5,5),1);
        Console.WriteLine("{0}", p.DistanceToPoint(new Point3(0,0,0)));

Instead I get the value of 0.115470053837925.
The only way I get a distance that makes sense is to create a plane using one of the three constructors below:
        Plane p = new Plane(new Vector3(0,0,1),1);
        Plane p = new Plane(new Vector3(0,1,0),1);
        Plane p = new Plane(new Vector3(1,0,0),1);

When using any of these constructors I get a distance of 1 from the origin.
What am I missing here?

The generic plane equation can be written as
$$
(p-p_0)\cdot \vec n = 0
$$
with $p = (x,y,z)$ In our case 
$$
\vec n = (5,5,5) =5\sqrt 3\left( \frac{1}{\sqrt 3}(1,1,1)\right)
$$
where $\frac{1}{\sqrt 3}(1,1,1) = \frac{\vec n}{||\vec n||}$ Choosing a plane at a distance $d$ from the origin means that
$$
p_0 = d\frac{\vec n}{||\vec n||} = 1\times \frac{1}{\sqrt 3}(1,1,1)
$$
so the sought plane is
$$
(p - 1\times \frac{1}{\sqrt 3}(1,1,1))\cdot (5,5,5) = 0
$$
or
$$
5x+5y+5z-5\sqrt 3 = 0
$$
0r
$$
x+y+z-\sqrt 3 = 0
$$
