Starting with an arbitrary matrix $A \in \mathbb{C}^{m\times m}$,
$$ A = \left[\begin{array}[cc] ** & * & * \\ * & * & * \\ * & * & * \end{array} \right] $$
where $*$ represents a any number. Produce
$$ A = \left[\begin{array}[cc] ** & * & 0\\ 0& * & * \\ 0& 0& * \end{array} \right] $$
multiplying $A$ by unitary matrices only.
The first 2 columns can be done with Householder reflectors, but I don't see how to get the zero in the (1,3) entry.
Product of unitary matrices is a unitary matrix, so we are to find, if possible, a unitary $U$ such that $$ U^*A=\begin{bmatrix}* & *&0\\\color{red}0&*&*\\\color{blue}0&\color{blue}0&*\end{bmatrix}. $$ Denote the columns of $U$ and $A$ as $$ U=[U_1\ U_2\ U_3],\quad A=[A_1\ A_2\ A_3]. $$ We have that $\color{blue}{U_3\bot A_1}$ and $\color{blue}{U_3\bot A_2}$. If $A_1$ and $A_2$ are not parallel then the only possibility for $U_3$ is along $A_1\times A_2$. Now $\color{red}{U_2\bot A_1}$ and $U_2\bot U_3$ gives one possible line again for $U_2$. It leaves only one direction for $U_1$, and the value of $U_1^*A_3$ cannot be chosen arbitrary. In general, it is not possible to get the third zero for a generic $3\times 3$ matrix $A$.