In a previous question (How can I show $\mathbf{e}_0\mathbf{e}_1\mathbf{e}_2\mathbf{e}_3=\sqrt{|g|}\gamma_0\gamma_1\gamma_2\gamma_3$) it was determined that
$$ \mathbf{e}_0 \wedge \mathbf{e}_1 \wedge \mathbf{e}_2 \wedge \mathbf{e}_3=\sqrt{|\det g|}\gamma_0 \wedge \gamma_1 \wedge \gamma_2\wedge \gamma_3 $$
Here, my question pertains to multivectors mixing blades of different grades, such as
$$ \mathbf{u}=v (\mathbf{e}_0 \wedge \mathbf{e}_1 \wedge \mathbf{e}_2 )+ b (\mathbf{e}_0 \wedge \mathbf{e}_1 \wedge \mathbf{e}_2 \wedge \mathbf{e}_3) = v\sqrt{|\det g[\mathbf{e}_0,\mathbf{e}_1,\mathbf{e}_2]|} \gamma_0 \wedge \gamma_1 \wedge \gamma_2+ b\sqrt{|\det g|}\gamma_0 \wedge \gamma_1 \wedge \gamma_2\wedge \gamma_3 $$
where $\sqrt{|\det g[\mathbf{e}_0,\mathbf{e}_1,\mathbf{e}_2]|}$ is of "dimension 3" and $\sqrt{|\det g|}$ is of "dimension 4".
I am trying to produce a multivector using a curvilinear basis that is invariant with respect to a change of coordinates. Specifically, I wish to multiple $\mathvf{v}$ defined as:
$$ \mathbf{v}=v \gamma_0 \wedge \gamma_1 \wedge \gamma_2+ b\gamma_0 \wedge \gamma_1 \wedge \gamma_2\wedge \gamma_3 $$
by $\sqrt{|\det g|}$ such that:
$$ \begin{align} \mathbf{v}'=\sqrt{|\det g|} \mathbf{v}&=v \sqrt{|\det g|}\gamma_0 \wedge \gamma_1 \wedge \gamma_2+ b \sqrt{|\det g|}\gamma_0 \wedge \gamma_1 \wedge \gamma_2\wedge \gamma_3 \end{align} $$
The term $b \sqrt{|\det g|}\gamma_0 \wedge \gamma_1 \wedge \gamma_2\wedge \gamma_3$ is equal to $ b \mathbf{e}_0 \wedge \mathbf{e}_1 \wedge \mathbf{e}_2\wedge \mathbf{e}_3$, but what is the (geometric) meaning/interpretation of the term $v \sqrt{|\det g|}\gamma_0 \wedge \gamma_1 \wedge \gamma_2$?
Note that it can also be written as:
$$ \frac{\sqrt{|\det g|}}{\sqrt{|\det g[\mathbf{e}_0,\mathbf{e}_1,\mathbf{e}_2]|}} \mathbf{e}_0 \wedge \mathbf{e}_1 \wedge \mathbf{e}_2 $$
The general question then, is what is the meaning of this term?
$$ \frac{\sqrt{|\det g|}}{\sqrt{|\det g[\mathbf{e}_0,\mathbf{e}_1,\mathbf{e}_2]|}} $$