I'm a student of electrical engineering. At our math class we were given a problem: given three points $A(1,2.3), \; B(-1,0,-1) \text{ and } C(5,4,3)$, find plane equation defined by these three points. Now, I began by solving this task myself. This is the procedure.
If norm of any plane is obtained via cross product of any two vectors proven to lay on the plane, then
$$ \vec{m} = \vec{AB} = (-2, -2, -4) \qquad \vec{n} = \vec{AC} = (4, 2, 0) $$
thus normal vector becomes $\vec{n} = \vec{m} \; \times \; \vec{n} = (8,-16,4)$. I actually double checked my answer using Symbolab. Next, using plane equation $\pi: \; ax+b y+cz=d$ I inserted normal vector components as coefficients and computed $d = 0$ (again, double checked that one as well).
Everywhere I looked at, this should be correct procedure, and although result is a bit confusing I believed my job was done. But solutions proved me wrong. Now this is where my comprehension ceases to exist; according to professor's notes, one could easily redifine vectors $\vec{m} \text{ and } \vec{n}$ as
$$ \vec{m} = (-2,-2,-4) = -2(1,1,2) \qquad \vec{n} = (4,2,0) = 2(2,1,0)$$
and use $(1,1,2) \text{ and } (2,1,0)$ to compute norm, which of course would now equal $\vec{n}=(-2,4,-1)$. The rest of the prodecure is obviously the same, just $d=3$ and not zero as I have computed.
Now, I know that this site is not meant for students to ask questions and anticipate full solution to their problems, so instead I'm just asking for a guideline. Can you actually 'shorten' vectors like this and continue operating with them?
Hint:
It is not important to use for the normal vector $\vec n=(8,-16,4)$ or $\vec n'=(-2,4,-1)$ since they are parallel vectors (and orthogonal to the plane).
Your mistake is in the value of $d$ in the equation of the plane. The value $d=0$ is for a plane passing through the origin, but you want a plane that passes through one of the given points $A,B,C$ so substitute the coordinates of one of these points in the equation $ax+by+cz=d$ and find $d$.