How to convert a plane (e.g. $4x - 3y + 6z = 12$) into parametric vector form? I can convert something in the 2nd dimension fine, but I'm having difficulty with something like $4x - 3y + 6z = 12$. Any help?
EDIT: Solve using only algebra, no matrices yet.
 A: Since there are three variables and one equation, you just denote the secondary variables as parameters, i.e. $y=s,z=t$ and then $x=\frac{12+3s-6t}{4}$. Then one parametric form is
$(\frac{12+3s-6t}{4},s,t)$. 
In the general case of a set of linear equations, it helps thinking of the equations that need parametrization as a system with more variables than equations. The key is to find how many secondary variables are there, and take them as parameters.
[edit:] You should be more precise in your formulation... 
The parametric vector form is very easy to obtain from the parametric vorm. Separate in three vectors separating $s,t$ and the constant term like this
$(\frac{12+3s-6t}{4},s,t)=(3,0,0)+s(\frac{3}{4},1,0)+t(\frac{-6}{4},0,1)$.
No parametric form is unique. If you replace $s,t$ by any linear combination of other two parameters, you get another parametric form, with different coefficients.
A: I cannot add a comment, therefore I have to post an answer. refering to the answer of Beni Bogosel and the question of meiro:
selecting $s=4v$ and $t=2u$ we get from the solution of Beni Bogosel
$$ x = (3, 0, 0) + u(-3, 0, 2) +  v(3, 4, 0) $$
there is a slightly different technic to get the parameter form 
$$\vec{x} = \vec{p} + s*\vec{a} + t*\vec{b}$$
$\vec{p}$ is a point of the given plane that means that the coordinates of $\vec{p}$ must satisfy the equation. for example if set the y and z coordinate of $\vec{p}$ to $0$ you get
$$ 4x-3*0+6*0=12$$
and therefor $x=3$ and $\vec{p} = 
\left( \begin{array}{c}
3\\ 
0\\
0
\end{array} \right)
$
. $ \vec{a} $ and $ \vec{b} $ are vectors parallel to the given plane. The coefficients 
$$ \vec{n}=\left( \begin{array}{c}
4\\ 
-3\\
6
\end{array} \right)
$$
of the equation represent a normal vector of the plane. Therefore $\vec{a}$ and $\vec{b}$ must be normal to $\vec{n}$. We can find a vector normal to a threedimensional vector $\vec{n}$ by taking  $\vec{n} $ setting one coordinate to $0$, interchanging the remaining coordinates and changing the sign of one of the interchanged coordinates (the inner product of this new vector with the original vector is now $0$, this means that they are normal). For example:
$$
\left( \begin{array}{c}
4\\ 
-3\\
6
\end{array} \right)
$$
set one coordinate to $0$, e.g. the $y$ coordinate
$$
\left( \begin{array}{c}
4\\ 
0\\
6
\end{array} \right)
$$
interchange the remaining coordinates
$$
\left( \begin{array}{c}
6\\ 
0\\
4\end{array} \right)
$$
and change the the sign of one of the interchanged coordinates, e.g. the sign of $6$. We get
$$
\left( \begin{array}{c}
-6\\ 
0\\
4\end{array} \right)
$$
which can be written as 
$$
2\left( \begin{array}{c}
-3\\ 
0\\
2\end{array} \right)
$$
We dont worry about length and orientation of the vector normal to $\vec{n}$ therefor we can use
$$
\vec{a}=\left( \begin{array}{c}
-3\\ 
0\\
2\end{array} \right)
$$
Taking $\vec{n}$ and setting the $z$-coordinate to $0$ and doing the analogous procedure we get the vector 
$$
\vec{b}=\left( \begin{array}{c}
3\\ 
4\\
0\end{array} \right)
$$
using the above equation for the vector $\vec{x}$ we get
$$

\vec{x}=

\left( \begin{array}{c}
3\\ 
0\\
0\end{array} \right)

+

s\left( \begin{array}{c}
-3\\ 
0\\
2\end{array} \right)

+

t\left( \begin{array}{c}
3\\ 
4\\
0\end{array} \right)
$$
A: Desribe one of the variables as composition of other two: x = (12 + 3y - 6z) / 4, and that gives you the parametric form of ( (12 + 3y - 6z) / 4, y, z)
