Find the general equation formed by three points

Question

I have to find the general equation defined by these 3 points:

What I have tried so far:

Unfortunately, I was told that my solution is wrong and I am not sure how else to solve/approach this problem. Any help is appreciated.

Answer 1

Hint: Write $$\vec{x}=\vec{OA}+s\vec{AB}+t\vec{AC}$$ where $s,t$ are real numbers.

Answer 2

Assuming we're after the equation of the plane $ABC$, you're on the right direction.

You already found a normal vector of that plane, though missed a sign, it should be $n=(-12,\ {\pmb +}12,\ -6)$, so the equation will indeed contain $-12x+12y-6y$ on one side, but the other side, which is going to be a constant, is still to be found.
For that, simply put the coordinates of a point on the $ABC$ plane (say, either $A$, $B$ or $C$) into this expression, and that will give you the constant.
[If they happen not to match for any choice of $A$, $B$ or $C$, that would mean you made a mistake in your previous calculations.]

So, now it's $$-12x+12y-6y\ =\ -12\cdot{\bf1}+12\cdot{\bf1}-6\cdot{\bf2}\ =\ -12$$ or, equivalently, $$2x-2y+z=2$$ You can verify that each of $A,B,C$ indeed satisfies it.

Answer 3

Hint:

Solve

$$\begin{cases} 1\cdot a+1\cdot b+2\cdot c+d=0, \\3\cdot a+1\cdot b-2\cdot c+d=0, \\0\cdot a-2\cdot b-2\cdot c+d=0. \end{cases}$$

You can set one of the coefficients arbitrarily and solve for the others.