Find a basis for all vectors perpendicular to $x-2y+3z=0$. I am looking at the following question:
Find a basis for all vectors perpendicular to $x-2y+3z=0$.
Clearly $<1,-2,3>$ is a vector that must be in the basis or a scalar multiple of $<1,-2,3>$ must be in the basis. But the solution given says that this vector is enough to describe a basis for all orthogonal vectors to the plane.
However, my initial intuition when solving this problem told me that this was not enough to describe the basis of all perpendicular vectors. My understanding is that any basis of a vector space is a minimal set of vectors whose linear combinations span the space.
My answer was that we would need to take a basis of the plane, say, $<2,1,0>$ and $<-3,0,1>$ and then combine that with $<1,-2,3>$. The reason for this is that a perpendicular vector need not be fixed. That is, we can move anywhere on the plane, and then move in the orthogonal dimension to that plane, by taking any scalar multiple of $<1,-2,3>$ and adding it to a linear combination of $<2,1,0>$ and $<-3,0,1>$. How does the basis $<1,-2,3>$ capture the orthogonal vector $<2,1,0>+<-3,0,1>+<1,-2,3>=<0,-1,4>$? Perhaps my plane/vector geometry is off. Also, my answer seems to suggest that the basis for the set of perpendicular vectors is also a basis for $\mathbb{R}^3$, which also seems odd.

The equation $$x-2y+3z=0$$ is the equation of a plain which is a two dimensional subspace in the three dimensional vector space of $\mathbb{R}^3.$
The set of vectors perpendicular to this plain constitute a one dimensional vector space which is just a line, spanned by any non-zero vector on that line. 
Since $<1,-2,3>$ is a non-zero vector on that line, the line will be spanned by this vector.
Thus a basis for your line consists of just that vector.  
