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:
$$ 7x - 5y - 4z = 6 $$

A plane can be defined by its normal $n$ and a point $p$ of the plane. The equation of the plane  is, for any point $x$ in the plane, $x\cdot n=d$, where $d=n\cdot p$.
If we have three points in the plane, $a,b,c$ then $a.n=b.n=c.n=d$. So we can solve for $n$ by forming a matrix $$Q=\begin{pmatrix}a&b&c\end{pmatrix}^T$$
and the three equations $$Q n=\begin{pmatrix}d&d&d\end{pmatrix}^T$$ must all be true if $a,b,$ and $c$ are in the plane, then we can get $n$ by inverting the matrix, $n=A^{-1}\begin{pmatrix}d&d&d\end{pmatrix}^T$.
In the terminology of your answer, $n\equiv{1\over d}(A,B,C)^T$, so what you did was to find the normal to the plane by inverting the above matrix.
This method works except for the case that the plane goes through the origin, where $d=0$ and we cannot use the inverse.
