 Solution

Assume $\mathbf{a}\ne 0$ and allow all $\mathbf{R}\ne 0$. Put $\hat{\mathbf{a}}=\mathbf{a}/|\mathbf{a}|$ and $\mathbf{u}=\mathbf{R}/|\mathbf{R}|$. Every resulting vector is perpendicular to $\mathbf{a}$ and has $|\mathbf{r}|=|c|\sin \theta \le |c|$. Conversely, for $c\ne 0$ and any $\mathbf{q}=\mathbf{r}/c$ perpendicular to $\hat{\mathbf{a}}$ with $|\mathbf{q}|\le 1$, chooseThe vector triple product gives $\mathbf{u}\times \hat{\mathbf{a}}=\mathbf{q}$, and $|\mathbf{u}|=1$. Thus every point of the closed disk centered at the origin, perpendicular to $\mathbf{a}$, of radius $|c|$ is attained. The boundary circle alone would omit allowed directions of $\mathbf{R}$. For $c=0$ the locus is the origin; for $\mathbf{a}=0$ the original expression is undefined.