I have been trying to solve for $\vec{r}$:
$\vec{a}\times \vec{r} + \lambda \vec{r} = \vec{c}$
And I thought of using a triple scalar product, I dot both sides with $\vec{a}$:
$\vec{a} \cdot (\vec{a}\times \vec{r}) + \lambda \vec{a} \cdot \vec{r} = \vec{a} \cdot \vec{c}$
And now since the triple scalar product is cyclic $\vec{a} \cdot (\vec{a}\times \vec{r}) = \vec{r} \cdot (\vec{a}\times \vec{a}) = 0$, so the main equation becomes:
$\lambda \vec{a} \cdot \vec{r} = \vec{a} \cdot \vec{c}$
And I do not know how to proceed from here. I cannot use the dot product again, $\vec{a}$ cannot be canceled out. I am not sure whether this approach is right in isolating the $\vec{r}$ but it seems closer to the answer.
Write $\vec{a}=a\hat{a}$ so that $\hat{a}$ is the unit vector in the direction of $\vec{a}$ and similarly $\vec{v}=v\hat{v}$. We have the equation $$av(\hat{a}\times\hat{v})+(\lambda v)\hat{v}=\vec{c}.$$ Notice that geometrically, $\hat{a}\times\hat{v}$ is a unit vector in a direction perpendicular to $\hat{v}$. Hence, $\hat{a}\times\hat{v}$ and $\hat{v}$ span a two-dimensional subspace of $\mathbb R^3$ (i.e. a plane). In order for a choice of $\vec{v}$ to work, $\vec{c}$ has to lie in this subspace, which translates to being orthogonal to the normal of the plane. Hence: $$\vec{c}\cdot\left( (\hat{a}\times\hat{v})\times\hat{v} \right)=0.$$ Once solving this equation for $\hat{v}$, we will explicitly know $\hat{a}\times\hat{v}$, and then the problem just becomes one of resolving $\vec{c}$ in terms of these orthogonal basis vectors. This can be done e.g. by just inverting a $2\times 2$ matrix.