Finding the equation of a plane given three points Below is a problem I did from a Calculus text book. My answer matches the
back of the book and I believe my answer is right. However, the method
I used is something I made up. That is, it is not the method described
in the text book.
Is my method correct?
Problem:
Find the plane through the points $(1,1,-1)$, $(2,0,2)$ and $(0,-2,1)$.
Answer:
The general form of a plane is:
$$ Ax + By + Cz = D$$
Sometimes the following constrain is added:
$$ A^2 + B^2 + C^2 = 1$$
By inspection, we can see this plane is not parallel to the x-axis, the y-axis or the z-axis. Hence,
we can assume that the plane is of the form:
$$ Ax + By + Cz = 1 $$
Now we setup the following system of linear equations.
\begin{align*}
A + B - C &= 1 \\
2A + 2C &= 1 \\
-2B + C &= 1 \\
\end{align*}
To solve this system of equations, we get rid of $A$ and $B$ in the first equation.
\begin{align*}
2A &= 1 - 2C \\
A &= \frac{ 1 - 2C }{2} \\
-2B &= 1 - C \\
B &= \frac{ C - 1 }{2} \\
\left( \frac{ 1 - 2C }{2} \right) + \left( \frac{ C - 1 }{2} \right)  - C &= 1 \\
1 - 2C + C - 1 - 2C &= 2 \\
- 2C + C - 2C &= 2 \\
-3C &= 2 \\
C &= -\frac{2}{3} \\
B &= \frac{ -\frac{2}{3} - 1 }{2} = -\frac{2}{6} - \frac{1}{2} \\
B &= -\frac{5}{6} \\
A &= \frac{ 1 - 2\left(  -\frac{2}{3} \right)  }{2} = \dfrac{1 + \dfrac{4}{3} }{2} \\
A &= \dfrac{7}{6}
\end{align*}
Hence the equation is:
$$ \left( \dfrac{7}{6} \right) A + \left( -\frac{5}{6} \right) B + \left(  -\frac{2}{3} \right) C = 1  $$
Clearing the fraction, we get the final answer of:
$$ 7A - 5B - 4C = 6 $$
As pointed out by Paul, the correct answer is:
$$ 7x - 5y - 4z = 6 $$

You can also use standard solution which is the equation of a plane passing through a given point and a line. Suppose a plane passes through point M(x_0, y_0, z_0) and a line with following equation:
$\frac {x-x_1}{l}=\frac {y-y_1}{m}=\frac {z-z_1}{n}$
Then the equation of plane is:
$\begin {vmatrix}x-x_0&y-y_0& z-z_0\\x_1-x_0&y_1-y_0&z_1-z_0\\l&m&n\end {vmatrix}=0$
Now take for example points A an B; the components of line AB are:
$l=x_B-x_A=2-1=1$
$m=y_B-y_A=0-1=-1$
$n=z_B-z_A=2+1=3$
Take for example point $A (x_1=1, y_1=1, z_1=-1)$ and point $C(x_0=0, y_0=-2, z_0=1)$ and construct the determinant:
$\begin {vmatrix}x-0&y+2& z-1\\1-0&1+2&-1-1\\1&-1&3\end {vmatrix}=0$
Now solve this equation you finally get:
$$7x-5y-4z=6$$
