How to determine if point P(vector) lies on lin L? (1) Determine whether the point P lies on the line l.
(2) If P is not in the line l, find the plane that contains l and P.
The line is represented like this: x = 1 + 2t, y = 2 - 3t, z = 3 - t. 
With any random points being P(x,y,z).
What would determine whether the points plugged in the equation fall on the line or not?
I've spent hours trying to figure this out but I cannot.
 A: (1) Given point $P(x,y,z)$, substitute those values into the equations for the line:
$$x=1+2t,\quad y=2-3t,\quad z=3-t$$
Those are three equations in one unknown (since you know $x,y,z$). If you find a value of $t$ that satisfies all three equations, $P$ is on the line. If that is not possible, $P$ is not on the line.
There are formulas I could give you, but they would not generalize easily to all lines. This method is quite fast enough.
(2) With your line, substituting $t=0$ shows that the point $A(1,2,3)$ is on the line. The direction vector for the line comes from the coefficients of $t$ in your parameterization and is $(2,-3,-1)$. If $P(x,y,z)$ is not on the line, the vector from point $A$ to point $X$, namely $(x-2,y+3,z+1)$, is another vector in the plane.
Take the vector cross product of $(2,-3,-1)$ and $(x-2,y+3,z+1)$ and get a vector $(a,b,c)$. That gives us the coefficients for the equation that defines your plane, namely
$$ax+by+cz=a+2b+3c$$
since substituting the point $A(1,2,3)$ into the left-hand side of that equation gives the value on the right-hand side.
