With research, I've mainly found that rowspace is the orthogonal complement of the nullspace, and the only vector that belongs to both spaces at the same time is {0}.
In the linear algebra subject that I'm doing so far, we have not learnt about orthogonality, and we're only expected to use the materials we've been given up until this point of the subject. More specifically, this is the question that we are to answer:
What is the proof that the condition given in (1b) is not possible?
Best Answer
In my opinion, the question is unfairly ambiguous. The rowspace of an $\ m\times n\ $ matrix comprises $\ 1\times n\ $ matrices (i.e. row vectors), so a $\ 3\times 1\ $ column vector such as $\ \mathbf{v}=\begin{bmatrix}1\\2\\3\end{bmatrix}\ $ cannot belong to it.
If, however, you interpret the question as instead asking for a matrix $\ \mathbf{A}\ $ such that:
$ \hspace{3em}\mathbf{v}^{\color{red}\top} \in\text{Row }\mathbf{A} $ and $\ \mathbf{v} \in\text{Nul }\mathbf{A}\ $ at the same time
then you can show this is impossible because it leads to a contradiction: \begin{align} \mathbf{v} \in\text{Nul }\mathbf{A}&\Rightarrow\mathbf{Av}=0\ \ \text{ and}\\ \mathbf{v}^\top \in\text{Row }\mathbf{A} &\Rightarrow \mathbf{v}^\top = \mathbf{x}^\top\mathbf{A}\ \ \text{ for some row vector }\mathbf{x}^\top\\ &\Rightarrow \mathbf{v}^\top \mathbf{v}= \mathbf{x}^\top\mathbf{A} \mathbf{v}=0\ , \end{align} which contradicts the fact that $\ \mathbf{v}^\top \mathbf{v}= 1^2+2^2+3^2=14\ $.