How can I find null space of orthonormal basis if $\operatorname{Null}(A)=0$? The question is

Find orthonormal bases for $R(A)$ and $\operatorname{Null}(A)$ where $$A=\left[\begin{matrix}
1 & 4 & 0\\
-2 & -3 & 1\\
0 & 0 & 2
\end{matrix}\right].$$

I found orthonormal basis for $R(A)$ and $\operatorname{Null}(A) = \{0\}$
If $\operatorname{Null}(A)=0$, how can I find orthonormal basis for $\operatorname{Null}(A)$?
 A: First of all, what is a basis for the subspace $\{0\}$? Well, it has to be a set $S$ of vectors in $\mathbb{R}^3$ which is linearly independent and has $\operatorname{span}(S) = \{0\}$. 
If $v \in \mathbb{R}^3\setminus\{0\}$ and $v \in S$, then $\{0\} \neq \operatorname{span}\{v\} \subseteq \operatorname{span}(S)$, so $\operatorname{span}(S) \neq \{0\}$. Therefore, $S$ cannot contain any non-zero vectors. 
What about the zero vector? If $S = \{0\}$ then $\operatorname{span}(S) = \{0\}$ but $S$ is not linearly independent as $c.0 = 0$ has non-trivial solutions. 
Therefore, $S$ cannot contain any non-zero vector or the zero vector so it must contain no vectors, i.e. $S = \emptyset = \{\}$. Note, by convention we set $\operatorname{span}\emptyset = \{0\}$ so that this is consistent. This also matches up with geometric intuition, because now $\dim\{0\} = |\emptyset| = 0$.
Recall that an orthonormal basis for a subspace is a basis in which every vector has length one, and the vectors are pairwise orthogonal. The conditions on length and orthogonality are trivially satisfied by $\emptyset$ because it has no elements which violate the conditions. This is known as a vacuous truth.
