TITLE: Find the equation of the plane knowing that it passes through 3 points
QUESTION [5 upvotes]: 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??

REPLY [2 votes]: We can do with some geometry and linear algebra.
We know area of of a triangle with three vertices is proportional to determinant :
\begin{vmatrix}
x_1 & y_1 & z_1\\
x_2 & y_2 & z_2\\
x_3 & y_3 & z_3\\
\end{vmatrix}
The determinant will vanish if the points are collinear.
It is derived by setting cross product to zero. Now let us pull in our general point.
Let one point be the general point P (x,y). I leave it to you to draw simple difference vector diagram of triangle where vector P is subtracted and other modified. When the area is zero, vector lines are in the same direction leaving zero enclosed area, so it is a situation that the general straight must now pass through plane determined by three points.
\begin{equation}
\begin{vmatrix}
x-x_1 & y-y_1 & z-z_1\\
x_2-x_1 & y_2-y_1 & z_2-z_1\\
x_3-x_1 & y_3-y_1 & z_3-z_1\\
\end{vmatrix}
=0
\end{equation}