Get the equation of a plane I need to get the equation of a plane in space. I didn't know how to do this so I looked it up and came across this:

Points: A(1,0,1) B(2,2,0) and C(3,1,4)
Direction: AB:(1,2,-1) and BC:(2,1,3)
/ x = 1 + r.1   + s.2
| y = 0 + r.2   + s.1
\ z = 1 + r.(-1)+ s.3

This far I understand everything, but I don't understand how to solve the following equation and how the author set up this equation:

|x-1    y       z-1 |
| 1     2        -1 |  =  0  <=> 7x - 5y - 3z - 4 = 0
| 2     1        3  |

Could anyone tell me how the author solved that?
 A: The determinant of a $3\times 3$ matrix tells you the volume of the corresponding parallelepiped formed by the three row vectors.
You can imagine a plane spanned by two vectors as being the set of vectors which form a volume $0$ solid. 
We have the two direction vectors relative to $A$ and so a vector $(x,y,z)$ will lie in the plane if and only if the volume formed by $(x-1,y,z-1), (1,2,-1), (2,1,3)$ is $0$.
Computing the determinant gives the answer you gave above.
A: The vector equation of a plane is $n\cdot (X - a) = 0$  where $n$ is a vector normal to the plane, $a$ is any point on the plane and $X$ is the general point on the plane, the point such that the equation holds if and only if $X$ is on the plane.  If you find a normal and point on the plane and expand that dot product your equation will be of the form $ax + by + cz = d$ where $a,b,c,d$ are real numbers and $(x,y,z) = X$.  To see this, dot product distributes over vector addition, so:
$$
n \cdot (X - a) = n\cdot X - n\cdot a = \\
n\cdot (x, y, z) - n\cdot a = 0
$$
So in the standard form $ax + by + cz = d$, $(a,b,c) = n$ and $d = n\cdot a$.  
If the author was using determinant somewhere it was probably to calculate a normal vector to the plane.  There's a formula for finding a vector $n$ perpendicular to two other vectors $X$, $Y$ called the cross product (in 3D).  $X \times Y$ ("$X$ cross $Y$") equals a vector perpendicular to both $X$ and $Y$.  As you can imagine there are two such vectors so the usual convention is to define $X \times Y$ by the right-hand-rule.  Point your right hand fingers in $X$'s direction and curl them along the shortest angle between $X$ and $Y$, until your fingers point in $Y$'s direction.  Your thumb then points in the direction of $X \times Y$.  A formula for calculating this given $X = (x_1, x_2, x_3), Y = (y_1, y_2, y_3)$ is
$$
X \times Y = \begin{vmatrix}
 \hat{x}& \hat{y}  & \hat{z} \\ 
 x_1 & x_2 & x_3\\ 
 y_1 & y_2  & y_3
\end{vmatrix}
$$
This is the matrix determinant formula for cross product.  Calculating the determinant for a $3\times 3$ matrix is relatively easy, then as you go up in matrix size the formula becomes more involved.
So the above formula expands to :
$$
(x_2 y_3 - x_3 y_2)\hat{x} - (x_1 y_3 - x_3 y_1) \hat{y} + \dots
$$
where $\hat{x}$ is your unit vector in the $x$-axis direction.  So you can rewrite the formula in parenthesis coordinate form as
$$(x_2 y_3 - x_3 y_2, -(x_1 y_3 - x_3 y_1), \dots )$$
I like to remember this formula in "index notation" as $(23 - 32, 31 - 13, 12 - 21)$.  But it's even easier to remember the "index 3-cycle" $123$.  To calculate coordinate $1$ it's $23 - 32$, then continue through the 3-cycle and when you get to the end, wrap around, so coordinate $2$ is $(31 - 13)$, and so on...
A: $\vec{n}
\equiv \left(\vec{B} - \vec{A}\right)\times\left(\vec{C} - \vec{A}\right)
\quad$. $\vec{n}$ is perpendicular to the plane $P:\ \vec{n} \perp P\quad$. $P$: set of plane points. Given a point $\vec{r} \in P$, it  satisfies
$$
\color{#ff0000}{\large\left(\vec{r} - \vec{A}\right)\cdot\vec{n} = 0}
$$
