I have a doubt about multivectors and linear independence.
Let $w_{1}$, $w_{2}$, $v_{1}$, $v_{2}$, $v_{3}$ be elements of some $n$-dimensional vector space, and assume that $v_{1} \wedge v_{2} \wedge v_{3} \neq 0$. Clearly, if both $w_{1}$ and $w_{2}$ are in the span of $v_{1}$, $v_{2}$, $v_{3}$, then for any $\mu \in \mathbb{R}$ there is $\lambda \in \mathbb{R}$ such that \begin{equation}\tag{1}\label{eq1} \left(v_{1}+ \mu w_{1}\right)\wedge \left(v_{2}+ \mu w_{2}\right)\wedge v_{3} = \lambda \left(v_{1} \wedge v_{2}\wedge v_{3}\right). \end{equation}
I am interested in the converse of this statement. More precisely, is it true that, if for some $\mu \in \mathbb{R}$ there exists $\lambda \neq 0$ such that equation \eqref{eq1} holds, then each of $w_{1}$ and $w_{2}$ is in the span of $v_{1}$, $v_{2}$, $v_{3}$?
EDIT: When $\lambda\neq 0$, equation \eqref{eq1} expresses the condition that the span of the vectors $v_{1}+ \mu w_{1}$, $v_{2}+ \mu w_{2}$, $v_{3}$ is equal to that of $v_{1}$, $v_{2}$, $v_{3}$.
The claim is false, take $\mu=1$, $w_1=v_2+v_3+v_4$ and $w_2= v_1+v_3+v_4$, where $v_4\not\in \text{span}(v_1,v_2,v_3)$. Then, $$(v_1+\mu w_1)\wedge (v_2+\mu w_2)\wedge v_3=(v_1+v_2+v_3+v_4)\wedge (v_1+v_2+v_3+v_4)\wedge v_3=0=0\cdot(v_1\wedge v_2\wedge v_3).$$