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How it comes that |D| is the distance of a plane to the origin?

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    do you mean the Hessian Normalform? – 2017-01-28
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    see here http://mathworld.wolfram.com/Point-PlaneDistance.html – 2017-01-28
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    @Dr.SonnhardGraubner I'm unsure, the book says: `point normal plane equation`. – 2017-01-28
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    @Dr.SonnhardGraubner I think the variable from http://mathworld.wolfram.com/Point-PlaneDistance.html `D` isn't the same as what the book refers as `D`. I've added additional pictures from the book to clarify it. – 2017-01-28

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For the plane $$ax+by+cz=D$$ the distance from the origin to the plane is $$d=\left|\frac{D}{\sqrt{a^2+b^2+c^2}}\right|$$

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As distance can never be negative as in your case -4. So they take magnitude of D. So D changes to |D|.

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$n$ is the normal vector whose direction is perpendicular to the place, i.e. the same direction in which we measure distance. The magnitude of $n$ is not really important here. All we need is the magnitude of the projection of $p$ on $n$ which is given by the absolute value of the scalar product of $p$ and $n$, which is $$p^{\prime}n=(4,0,0)^{\prime}(1,1,1)=4 \times 1 + 0 \times 1 + 0 \times 1=4$$ Hence the required distance. The picture presented there is helpful in understanding why the projection leads to the distance. Hope it helps.

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    I think my confusion is at the dot product of `p'n`, in trigonometric for the projection of `p onto n` would be `||po|| * cos(theta)` and yes that would be equal to `D`, but dot product is `||po|| * cos(theta) * ||n||`, shouldn't `n` be a unit vector? – 2017-01-28
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    Well the normal seems to be unit length. – 2017-01-28