I came across a result which seems to be true (via a good number of computer calculations for randomly generated examples) but that I do not know how to prove:
Let $V$ be an orthogonal real $n \times n$ matrix and $B$ be any $n \times n-1$ real matrix. Then it appears that $\forall 1\leq i\leq n$:
$$\det\begin{bmatrix}V^TB & \pmb{e_i}\end{bmatrix} ^ 2 = \det\begin{bmatrix}B & \pmb{v_i}\end{bmatrix} ^ 2$$
where $\pmb{e_i}$ is the standard basis vector, $\pmb{v_i}$ is the $i$-th column of $V$, and the matrices in the determinant on the LHS and RHS are each $n \times n$ and formed by appending the $n \times n-1$ matrix on the left with the $n \times 1$ vector on the right.
I can demonstrate it clumsily for $n=2$ and thought about setting out for an inductive proof, but I imagine there must be a more direct, illuminating way - would appreciate even the smallest hint or general direction on what that might be!
(Apologies for the long title, let me know if using notation is preferable to words.)
Hint: Note that $$ V^T \pmatrix{B&v_i} = \pmatrix{V^TB & V^Tv_i} = \pmatrix{V^TB & e_i}. $$