How about this approach:
Orthogonal projection can be thought as the (smallest) distance from this point to the plane.
Assume that the projection is (a,b,c). Then you have distance d as:
d^2 = (a-3.5)^2+(b-1.5)^2+(c+1.5)^2
Because (a,b,c) is a point on the plane, so you also have
Then you can combine the above two, and get
d^2 = (a-3.5)^2+(b-1.5)^2+(b-a+4.5)^2
Now you need to minimize d basically. This will give you a, b, and c.