write the vector equation of the plane with equation $2x-3y+4z=12$ I don't know where to start with this question. Usually I'm given an initial point and a vector to find the vector equation. But here I'm only given an equation. How can I find the vector equation from an equation?
 A: A normal vector would be $\vec n = (2,-3,4)$. A point on the plane would be: $(0,0,3)$. Thus:
$$\vec n \cdot (x,y,z-3)= 0$$
is a vector equation.
A: Hint: use the given equation to calculate some points that lie in the plane (for example just put in $x=0,y=0$ and solve for $z$, then repeat with $x=0,z=0$ and $y=0,z=0$).
Edit: I just realised that you didn't specify which vector equation you mean. If you're going for an equation in normal form, BolzWeir's way is much simpler.
A: Find the point and the vectors.
To find the point just give $x$ and $y$ (for example) the values that you prefer in the equation of the plane, and solve for $z$. Call this point $P$.
For the two vectors, repeat twice the former process to obtain two more points $Q$ and $R$. You can use the vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$. You must verify that $P$, $Q$ and $R$ are not in a line. If they are, you must find another point.
Alternatively, when you have $\overrightarrow{PQ}$ you can find $\overrightarrow{PR}$ letting $\vec n$ a vector normal to the plane, that is $\vec n=(2,-3,4)$, and calulating $\overrightarrow{PR}=\vec n\times \overrightarrow{PQ}$.
A: There's a lot of good methods already posted, but I'm missing an important interpretation here. A plane is a "level-set" of a linear function, i.e. the set of inputs which give a constant value to the function. That means that the function 
$$2x-3y+4z$$ is constant along the plane. Level-sets are very important in for instance geometry applications. Planes, circles, parabola, ellipses and many other geometrical shapes can be described as level sets.
