**Part 1**

The problem needs a bit more specification - “in the same plane as the predetermined vector” doesn’t define a unique plane. For instance, the x=3 plane passes through it, as do the y=1 and z=2 planes. Generally, you need three points to define a plane, or a normal to the plane and one point (usually, a point on that normal vector), or some other definition with enough constraints in it.

Taking the diagram as indicative, though, what it /looks/ like is that the plane shown is defined as including also the y-axis (assuming y-axis is marked green, as in Unity). In which case, it’s the plane 2x=3z, which can also be defined as having the normal vector [2, 0, -3] (errr… I think…) and passing through the origin. We could define the plane shown with these two data:

`Normal vector N = [2, 0, -3]`

`Point in plane P = [0, 0, 0]`

Doesn’t matter if that’s actually your plane, the point is you need it in some standard form.

**Part 2**

The problem needs a bit more specification, though, again - “in the same plane as the predetermined vector” doesn’t define a unique point in a given plane. The diagram seems to be showing projection along the blue axis (z?), but the vector (2, ?, -2) suggests it’s a projection in y. So you’ll need to specify more about the result than just it being in the plane.

A common extra bit of information is to say ‘the nearest point on the plane’, so I’ll go with that, but a projection along any axis, or indeed any direction, is also a thing you might want to do. The procedure for those is a bit different, but the tools are from the same toolbox.

**Part 3**

Assuming you have a normal vector, N, and any point in the plane P, you can project your ‘random vector’ X onto the nearest point on that plane by doing:

`X += Vector3.Project(P, N) - Vector3.Project(X, N)`

noting that you can precompute the first term since it’s the same for any X.

The fact that this is so neat underlines why people like to represent planes in N/P format.

Forgive any silly mistakes, I’m not testing this.