$\def\qty#1{\left( #1 \right)}$
$\def\R{\mathbb{R}}$
Start with the definition of the determinant of a matrix $\def\R{\mathbb{R}}$
$C \in \R^{3 \times 3}$:
$$
\det C := \epsilon^{ijk} {C^1}_i {C^2}_i {C^3}_i
$$
where, for example, ${C^1}_i$ is element $i$ of the first row of the matrix.
If $C$ is the product between two matrices $A,B$ then, since ${C^i}_j={A^i}_k{B^k}_j$, we can rewrite the formula for the determinant as,
$$
\det C = \epsilon^{ijk} {A^1}_p{B^p}_i {A^2}_q{B^q}_j {A^3}_r{B^r}_k
\tag{1} \label{1}
$$
It is also the case that,
$$
\epsilon^{ijk} {B^p}_i {B^q}_j {B^r}_k
=\det B \; e^{pqr}
$$
and so,
$$
\det C = \det B \; e^{pqr}{A^1}_p {A^2}_q {A^3}_r
= \det B \; {A^1}_p \qty{e^{pqr} {A^2}_q {A^3}_r}
= \det B \; {A^1}_p \qty{A^2 \times A^3}^p \tag{2} \label{2}
$$
We can rewrite ($\ref{1}$) as,
$$
\det C = \epsilon^{ijk} {A^1}_p{B^p}_i {A^2}_q{B^q}_i {A^3}_r{B^r}_i
=
\qty{{A^1}_p{B^p}_i} \qty{\epsilon^{ijk}{A^2}_q{B^q}_j {A^3}_r{B^r}_k}
= \qty{{A^1}_p{B^p}_i} \qty{C^2 \times C^3}^i
$$
Now set $B=Q^T$ the transpose of an orthogonal tensor $Q$ which has the property,
$$
\qty{a_p {Q_i}^p} \qty{{Q^i}_r b^r}
= a_p \qty{{Q_i}^p {Q^i}_r} b^r
= a_p \delta^p_r b^r = a_p b^p
$$
(${Q_i}^p {Q^i}_r={\qty{Q^TQ}^p}_r={I^p}_r=\delta^p_r$). We can use this property to rewrite,
$$
{A^1}_i \qty{A^2 \times A^3}^i
=
\qty{{A^1}_p {Q_i}^p} \qty{{Q^i}_r \qty{A^2 \times A^3}^r}
\tag{3} \label{3}
$$
Combining ($\ref{1}$), ($\ref{2}$) and ($\ref{3}$),
$$
\det Q^T \; \qty{{A^1}_p {Q_i}^p} \qty{{Q^i}_r \qty{A^2 \times A^3}^r}
=
\qty{{A^1}_p{Q_i}^p} \qty{C^2 \times C^3}^i
$$
So we must have,
$$
\qty{C^2 \times C^3}^i = \det Q \qty{{Q^i}_r \qty{A^2 \times A^3}^r}
$$
($\det Q^T=\det Q$).
If ${A^2}_i=v_i$ and ${A^3}_i=w_i$ then ${C^2}_i={A^2}_p {Q_i}^p= v_p {Q_i}^p = (Qv)_i$ and ${C^3}_i={A^3}_p {Q_i}^p= w_p {Q_i}^p = (Qw)_i$. So the last equation is equivalent to,
$$
\qty{Qv}\times\qty{Qw}=\det Q \; \qty{Q \qty{v \times w}}
$$
