Suppose we have three non orthogonal vectors in $R^3$ as $\vec{a}, \vec{b}, \vec{c}$. The vector of $(\vec{b}\cdot \vec{c})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}$ is in the plane spanned by $\vec{a}$ and $\vec{b}$, and it is perpendicular to $\vec{c}$. Is there a similar geometric explanation for vector $$(\vec{a}\cdot \vec{b})\vec{c} + (\vec{b}\cdot \vec{c})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}$$?
EDIT: Answer by Muphrid indicates this is a normal vector (with dilation) of three consecutive reflections to normal vector $a, b, c$. Can this be carried out without using geometric products? Starting from a reflection as $x' = x - 2(x \cdot n)n$, it gets complicated for me.
Recall that any reflection map $\underline N$ across a plane perpendicular to a unit vector $n$ can be written as
$$\underline N(a) = -nan$$
Interpret the product $abc$ then as a composition of reflections: the vector part that you found is the normal vector to the net reflection plane that this composition results in.
Edit: actually, let $v$ be that vector and $T$ be the trivector part. Composing the reflections gives
$$-(v+T)x(v-T) = -vxv - Txv + vxT + TxT$$
The first term is a reflection-dilation. The second and third factor to $T(vx-xv) = 2T (v \wedge x) = 2(Tv) \cdot x$, giving a vector in the plane $Tv$ orthogonal to $x$. The last term is $T^2 x$, with $T^2 \leq 0$.
So there is still a net reflection plane, but it does not uniquely characterize the composition of reflections, as there are some other terms involved.