I have these 2 equations: $$\begin{align} (68-4i,44+12i,-38-2i)\mathbf x=0 \\ (-66i-18,-52i-10,42i+12)\mathbf x= 0 \end{align}$$
I need to find the span of vector $\mathbf x \in \Bbb C^3$. I'm supposed to find a vector which is orthogonal to both of the vectors above.
To find a vector orthogonal to both the vectors, you can apply the vector cross product to the two vectors as follows
$$u\times v=\left| \begin{array}{ccc} \mathbf{a} & \mathbf{b} & \mathbf{c}\\ \color{red}{68−4i} & \color{blue}{44+12i} & \color{green}{−38−2i} \\ \color{red}{−66i−18} & \color{blue}{−52i−10} & \color{green}{42i+12}\end{array} \right|\\= \left|\begin{array}{cc} \color{blue}{44+12i} & \color{green}{−38−2i} \\ \color{blue}{−52i−10} & \color{green}{42i+12}\end{array}\right|\mathbf{a}- \left|\begin{array}{cc} \color{red}{68−4i} & \color{green}{−38−2i} \\ \color{red}{−66i−18} & \color{green}{42i+12}\end{array}\right|\mathbf{b} +\left|\begin{array}{cc} \color{red}{68−4i} & \color{blue}{44+12i} \\ \color{red}{−66i−18} & \color{blue}{−52i−10}\end{array}\right|\mathbf{c}\\=(-252-4i,-432-264i,-888-376i)$$
Thus the span of vector $\mathbf{x}$ is $\color{blue}{k(-252-4i,-432-264i,-888-376i)}$, where $k$ is a complex scalar constant.