For convenience, let's first reproduce here the image given in the answer of [ViewMatrix from a Graphics3D](http://mathematica.stackexchange.com/a/3538/5478):

[![3D view explanation][1]][1]


> Where the BC segment is horizontal.

If the BC segment is supposed to be horizontal, then `ViewVertical` (refer to the image) must be orthogonal to BC.

    v1 = Normalize[c - b];
    v2 = Normalize[(a - b) - Projection[a - b, c - b]];

`v1` and `v2` are orthogonal to each other. `v1` is BC, horizontal by definition, and therefore `v2` is the view vertical.

We can also compute 

    v3 = Cross[v1, v2];

so that we have a complete, orthonormal, basis for the 3D space.

    a = {2, 5, 9};
    b = {3, -3, 2};
    c = {5, -4, -6};
    d = {-1, 7, 8};
    
    gr = Graphics3D[{
       Text["A", a + .3],
       Text["B", b + .3],
       Text["C", c - .3],
       Text["D", d + .3],
       Line[{a, b, c}],
       Blue, PointSize[.02],
       Point[{a, b, c}],
       Red, Point[d],
       Black, Opacity[0.2], InfinitePlane[{a, b, c}],
       Red, Thick, Opacity[1],
       InfiniteLine[{b, b + v1}],
       InfiniteLine[{b, b + v2}],
       InfiniteLine[{b, b + v3}]
       }, PlotRange -> 10]

![Mathematica graphics](https://i.sstatic.net/KBH8N.png)

> How can I set the viewpoint so that I have a view perpendicular with
> regard to plane formed by these three points ABC selected

Again referring to the image, we see that the line between the view center and the view point must be perpendicular to the plane spanned by `v1` and `v2`. But default the `ViewCenter` is in the middle of the plot range as in the image. For simplicity we choose a plot range that runs between -10 and 10 in all three directions, that way the view center is at (0,0,0). This means that the view point must lie along `v3`. Testing this:

    Show[gr, ViewPoint -> 10 v3, ViewVertical -> v2, Boxed -> False]

![Mathematica graphics](https://i.sstatic.net/vC7hy.png)

We see that this worked out well. The segment BC is horizontal, and only two axes are visible which means that we are looking into the plane at a perpendicular angle.



  [1]: https://i.sstatic.net/cemyD.png

> And what is the distance between the point D to the plane.

The distance is the orthogonal projection of the point onto the plane:

    d.v3 // N
    (* Out: 0.354459 *)