How can we determine the numbers $k_1$ and $k_2$ ? Suppose that the matrix $A=\begin{pmatrix}1+ci & w_1 \\ 2+i & z_2\end{pmatrix}$ with $c\in \mathbb{R}$ is hermitian, of order $1$ and the vector $(k_1, k_2)\in \mathbb{C}^2$ with $k_2$ positive real number, belongs to the orthogonal complement of the row space of $A$ and has norm $\sqrt{3}$. 
How can we determine the numbers $k_1$ and $k_2$ ? 
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We know that $A$ is hermitian, so it is equal to its own conjugate transpose. So we have $A=\overline{A^T}$, i.e. $$\begin{pmatrix}1+ci & w_1 \\ 2+i & z_2\end{pmatrix}=\begin{pmatrix}1-ci & 2-i \\ \overline{w_1} & \overline{z_2}\end{pmatrix}$$ From that we get that $c=0$, $w_1=2-i$ and $z_2=\overline{z_2}$ and so $z_2$ is real. 
Are the information correct so far? 
Then we have that $A$ has the order 1, that means that $A^1=I_2$, or not? 
Now we have to calculate the orthogonal complement of the row space of $A$, or not? How can we do that? 
The last information is that the vector $(k_1, k_2)$ has the norm $\sqrt{3}$ and so we get $\sqrt{k_1^2+k_2^2}=\sqrt{3}\Rightarrow k_1^2+k_2^2=3$, right? 

Just to put the information in all the comments together:
You may make use of the fact that the orthogonal complement of the row space of $A$ is the kernel of $A$ (that is easy to prove.)
Then, $z_2$ may be determined, keeping in mind that there is at least 1 non-zero vector in the kernel of $A$ (which vector?). Then find what the kernel is.
Finally, you may find one (or more maybe?) vectors $(k_1, k_2)$ in the kernel of $A$ which satisfies the requirements given for $k_2$ and $\|(k_1, k_2)\|$. 
