Generating a $n$-th dimensional vector orthogonal to $n-1$ linearly-independent vectors Let us have $n-1$ linearly independent vectors $\vec{v}_{1},\dots,\vec{v}_{n-1}\in\mathbb{R}^{n}$, define the vector $\vec{w}$ as follows:
$$\vec{w}=\begin{pmatrix}\det(\vec{v}_{1},\dots,\vec{v}_{n-1},\vec{e}_{1}) \\ \vdots \\ \det(\vec{v}_{1},\dots,\vec{v}_{n-1},\vec{e}_{n})\end{pmatrix}$$
This vector is apparantly orthogonal to all vectors $v_{i}$, $i\in\{1,\dots,n-1\}$; however I am not able to prove that this is the case and I don't understand the intuition behind it either, so I would be grateful if someone was able to prove this fact and perhaps allow me to grasp it more intuitively.

If you write out the dot product and expand each determinant along the last column, you'll see that $$\vec{v}_i\cdot \vec{w} = \det(\vec{v}_1, \vec{v}_2,\ldots, \vec{v}_{n-1},\vec{v}_i),$$ and since the right hand side is the determinant of a matrix with two equal columns, it is zero. I'm not sure if there is geometric intuition.
