Find the basis for all vectors perpendicular to plane $x-2y+3z=0$ in $\mathbb{R}^3$ This can be rewritten as $A\vec{x}=0$, where $A=(1, -2, 3), \vec{x}=(x, y, z)^T$.
I understand that one can find vectors perpendicular to this place by finding the basis for the null space, as the null space is perpendicular to the row space.
This basis is $(2, 1, 0), (-3, 0, 1)$. I believe that the space describing all perpendicular vectors can be written as $a(2, 1, 0)^T+b (-3, 0, 1)^T$? In order to find the basis for this space, we solve for the null space of 
$$
  B=
  \left[ {\begin{array}{cc}
   2, 1, 0 \\
   3, 0, 1 \\
  \end{array} } \right]
$$
This gives you $(1, -2, 3)^T$ which makes sense as this described the original plane. But I don't understand why we combine the two solutions for the null space and transpose them to find the basis for all vectors $a(2, 1, 0)^T+b (-3, 0, 1)^T$. Could somebody explain this to me? Thank you.
Note: This question is about the same problem but has a different misunderstanding.

Because we're working in a $3$-dimensional vector space and we have found $2$ linearly independent solutions $s_1$ ans $s_2$. Therefore, the space $S$ of all solutions must have dimension at least $2$. But it can't be $3$, since that would meant that $S=\mathbb{R}^3$, but it isn't ($(1\ \ -2\ \ 3)^T\notin S$). Therefore, $\dim S=2$ and $\{s_1,s_2\}$ is a basis of $S$.
