Let Ox, Oy and Oz be three semi-straight lines on which there are the points A, B and C so that OA=x, OB=2x and OC=3x. Find the geometric place of the projection of O onto the plane (ABC).
There are, I am afraid, problems with your English that make this hard to understand. "Semi straight lines" implies lines that are "almost" or "partially" straight, whatever that means! But I am pretty sure you mean "half line" (or perhaps a "ray") that is straight. Essentially what you have is a tetrahedron with vertices A, B, C, O and three of its six edges, AO, BO, CO have lengths of a, 2a, and 3a, respectively (I am using "a" rather than x because I want to reserve "x" for a coordinate). The thing I am not clear on is whether you intend the three "half lines" to be orthogonal, creating an orthogonal coordinate system. If that is the case then we can take O to be the origin, (0, 0, 0), A to be (a, 0, 0), B to be (0, 2a, 0), and C to be (0, 0, 3a). The plane determined by A, B, and C is x/a+ y/(2a)+ z/(3a)= 1 or 6x+ 3b+ 2z= 6a. A vector perpendicular to that plane is <6, 3, 2>. A line through the origin in the direction of that vector, so perpendicular to that plane, passing through (0, 0, 0) can be written in parametric equations x= 6t, y= 3t, z= 2t. That intersects the plane when 6(6t)+ 3(3t)+ 2(2t)= 49t= 6a or t= (6/49)t. That is the point x= 6(6/49)= 36/49, y= 3(6/49)= 18/49, z= 2(6/49)= 12/49. The point of projection of (0, 0, 0) onto the plane determined by those three points is (36/49, 18/49, 12/49).