Question

Projecting point on plane defined by normal distance form along direction

Answer 1

Multiply the unit normal vector by the distance, and subtract that vector from your point.

projected_point = point - dist*normal;

Edit with picture: I've modified your picture a bit. Red is v; v dot normal = length of blue and green (dist above). Blue is normal*dist. Green = blue * -1 : to find planar_xyz, start from point and add the green vector.

Answer 2

Answer:

The normal vector of a plane, perpendicular to the given line, coincides to the direction vector of the line, that is N = s = -3i + 5 j + 3k. Thus, the coordinates of the intersection or the projection A′ are, x = -3 · (-1) - 1 = 2, y = 5 · (-1) + 3 = -2 and z = 3 · (-1) + 2 = -1, A′(2, -2, -1).