Find the equation of the plane knowing that it passes through 3 points

I have to find the equation of the plane that passes through $(0, 0, 0), (4, 0, -2), (0, 8, -6)$.

I have done the following:

The equation of the plane is of the form $$ax+by+cz+d=0$$

Since the points $(0, 0, 0), (4, 0, -2), (0, 8, -6)$ are points o fthe plane, we have that $$d=0 \\ 4a-2c=0 \Rightarrow a=\frac{c}{2} \\ 8b-6c=0 \Rightarrow b=\frac{3c}{4}$$

So $$\frac{c}{2}x+\frac{3c}{4}y+cz=0 \Rightarrow \frac{1}{2}x+\frac{3}{4}y+z=0$$

Is it correct??

Is there also an other way to find the equation of the plane??

Maybe using the cross-product??

Hint: first you find the normal to the plane: $N = \begin{pmatrix} 4 \\ 0 \\ -2 \end{pmatrix} \times \begin{pmatrix} 0 \\ 8 \\-6 \end{pmatrix} = (a,b,c)$, then your plane is:

$a(x-0)+b(y-0)+c(z-0) = 0$

Since you mentioned the cross product, let's use it. If you have a plane $\Pi$, and $P,Q,R \in \Pi$, then the vectors $Q-P$ and $R-P$ lie in the plane. Hence $(Q-P)\times(R-P)$ is normal to the plane.

In your exercise, $(4,0,-2)$ and $(0,8,-6)$ lie in the plane. Then $(2,0,-1)$ and $(0,4,-3)$ lie in the plane too. So: $$\begin{vmatrix} {\bf i} & {\bf j} & {\bf k} \\ 2 & 0 &-1 \\ 0 & 4 & -3\end{vmatrix} = (4,6,8) $$is normal to the plane. Let's use $(2,3,4)$ instead. We have that the plane is $\Pi: 2x+3y+4z = 0$. (I'm already using that ${\bf 0}\in \Pi \implies d = 0$).