In $\mathbb{R}^3$, the vector space spanned by two vectors $\vec{v}_1, \vec{v}_2$ has a normal vector $$\vec{n}=\vec{v}_1\times \vec{v}_2$$ such that $$\vec{v}\in\text{span}(\vec{v}_1, \vec{v}_2)\iff \vec{n}\cdot \vec{v}=0$$
Question: Does this generalize to vectors in $\mathbb{C}^3$? It is fairly clear that $\vec{n}\cdot \vec{v}=0$ defines a vector space, even in $\mathbb{C}^3$, but it is not so clear that this space is still equivalent to the (complex) span of $\vec{v}_1$ and $\vec{v}_2$.
I think there is, perhaps, more going on than you think.
Let's suppose for a moment that you mean "complex dot product", which is unusual, but apparently what you are talking about, along with the definition of cross product that looks like
$$ \mathbf v\times \mathbf w=\left|\begin{matrix}\mathbf i & \mathbf j & \mathbf k \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3\end{matrix}\right| $$
Together they satisfy $$ \mathbf x\cdot(\mathbf v\times \mathbf w)=\left|\begin{matrix}x_1 & x_2 & x_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3\end{matrix}\right| $$
Then in your situation, where you are presumably assuming $\mathbf v=\vec{v_1}$ and $\mathbf w=\vec{v_2}$ are linearly independent, the dot product is zero iff the matrix on the right is singular, and that occurs exactly when the top row is in the span of the bottom two rows.
But the normal context for vector algebra with complex vectors uses the complex inner product, the one that is antilinear in a coordinate. What happens in that case? For starters, one really ought to adapt the cross product to suit the same identity above, and that is what is discussed in this post. In this context, you would be working with the ordinary complex inner product, and the "complex cross product". Since the same identity holds, the same conclusion holds.
