Finding the equation of a plane given three points where the solution does not have $z$ in it

Question

Below is a problem that I made up. I believe that my solution is correct but the answer seems to simple to me. Is my solution correct?
Problem:
Find the equation of the plane containing the points $(1,1,1)$, $(2,2,2)$ and $(2,2,3)$. Answer:

\begin{align*} P1 &= (1,1,1) \\ P2 &= (1,1,2) \\ P1 \times P2 &= \begin{vmatrix} i & j & k \\ 1 & 1 & 1 \\ 1 & 1 & 2 \\ \end{vmatrix} \\ P1 \times P2 &= (2 - 1 )i - (2-1)j + (1 - 1)k \\ P1 \times P2 &= i - j \\ (x-1) - (y-1) &= 0 \\ x - y &= 0 \end{align*} Partial check, is the point $(2,2,3)$ on the plane? $$ x - y = 2 - 2 = 0 $$. It is.

Answer 1

Your solution is indeed correct. I suppose that you will agree that the expression $x-y=0$ defines some plane. But all of your points belong to the plane $x-y=0$. Since three non-colinear points define a plane, then the plane that you are interested in has to be the plane $x-y=0$.

Answer 2

Your method is not a valid method; it worked in your case only because the plane passes through the origin. (You also have a typo. The first time you said $P_2 = (2,2,2)$. Note that if you'd used this point, the cross-product would have been $\vec 0$ to start with. Then what?) To convince yourself, try $P_1=(1,1,1)$, $P_2=(0,-1,3)$, $P_3=(2,3,4)$.

In general, to obtain the normal vector to the plane you must look at $(P_2-P_1)\times (P_3-P_1)$ or something analogous to that.