how to find orthogonal vectors of 3 Dimensional point If $A(3, 4, 5)$ and $B(7, 10, 12)$ are two end points of a line segment. I know that vector $V_1=B-A$  i.e $V_1=(4, 6, 7)$ then how to find the other two orthogonal vectors of this?  
 A: Three non-colinear points are needed to define a plane, but you only really have two here.  If you just want to find a couple of arbitrary vectors orthogonal vectors, just pick any third point, C, not colinear with A and B, then W = (A-C)×(B-C) (where × is the cross product) is orthogonal to the line segment connecting A and B since any plane containing A and B also contains the line segment connecting them.
If you need a third vector U orthogonal to V = A-B and to W, then just set U = V×W
The first step is probably easiest if you pick C as the origin (again if it's not colinear with A and B); this reduces to setting W = (A×B).
More information is available both at Wolfram and Wikipedia.
A: Use vector product:  V1 ^ V2, if they aren't collateral, it will give you an orthogonal vector.
A: There are not two unique orthogonal vectors, there are an infinity of them.
Consider V1 as the vector normal to a plane (in Euclidean space).  Any two orthogonal vectors lying in that surface will be orthogonal to V1.
Imagine a clock face.  Put the hands at 12 and 3.  You now have a set of 3 orthogonal vectors, V1, 12'o'clock, and 3'o'clock.  Rotate the clock face, and you get a different set.  Mirror-image the clock face, and you get another set.  And so on...
