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For example, there are the two planes as the following. \begin{eqnarray} \\C1\space :\space 2x+4y+3z=10 \space \space ,\space \space C2\space :\space 4x+8y+6z=-2 \end{eqnarray}

Why the vector normal line (n) can be (2, 4, 3) or (4, 8, 6) ?

I cannot understand. Please explain to me.

Thank you for your attention

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Only the direction matters. When you point towards some direction, it doesn't matter how long your finger is. – copper.hat Nov 25 '13 at 8:10

The trick about getting a good geometrical grasp of affine planes (that is, planes which do not necessarily go through the origin) is to realize that they are simply translations of a parallel algebraic plane (specifically, $P$ is affine $p\in P$ then $S=\{q-p|q\in P\}$ is algebraic.

With that in mind, given that a plane is parameterized by the equation $ax+by+cz=w$, you can interpret that as the set of all vectors $(x,y,z)$ such that $(x,y,z)\cdot(a,b,c)=w$. From here it follows that another affine plane is parallel to $(x,y,z)\cdot(a,b,c)=w$ iff it is of the form $(x,y,z)\cdot(a,b,c)=w'$ (one way to see this is to prove that if you choose differently one or more of the coefficients $a,b,c$ you can find a point which satisfies both equations).

Another insight is that an affine plane parameterized by $ax+by+cz=w$ is algebraic iff $w=0$. It should be clear why this is true.

Using this insight, you get that the unique algebraic plane parallel to $ax+by+cz=w$ is $ax+by+cz=0$.

However, if you assume that $n$ is orthogonal to a given plane it will be orthogonal to all parallel planes, so it suffices to find a vector orthogonal to the algebraic plane $ax+by+cz=0$.

However -- same as before -- saying that a vector $(x,y,z)$ is in the plane $ax+by+cz=0$ is equivalent to saying that $(x,y,z)\cdot(a,b,c)=0$ which is, in turn, equivalent to saying that $(x,y,z)\perp(a,b,c)$.

Hence, $(a,b,c)$ is normal to $ax+by+cz=0$ and by the above to any plane of the form $ax+by+cz=w$.

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