Finding the equation of the plane Passing through the points $(1,1,0)$, $(2,0,2)$, and $(0,3,3)$.
How do I know which components to subtract with which other component, in what order, to get it in ijk-form?
I know that the cross product of the two vectors will give me the normal vector. And from there I know how to get the equation of the plane.
\begin{align}&((2-1)i - (0-1)j + (2+0)k) \times ((0-1)i + (3-1)j + (3-0)k
\\&= (2i-j+2k) \times (-i+2j+3k) \end{align}
It's supposed to be $(2i-j+2k) \times (i-2j-3k)$
I'm just having troubles knowing which components to subtract.

An alternative approach to solve the problem:
A plane (not passing through the origin) can be described by
$$ax+by+cz = 1.$$
Use the three points to compute $a$, $b$, and $c$. Your final answer should be $$7x + 5y -z = 12.$$
