= Solution
Assume $\mathbf a\ne0$ and allow all $\mathbf R\ne0$. Put $\widehat{\mathbf a}=\mathbf a/|\mathbf a|$ 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 $c\ne0$ and any $\mathbf q=\mathbf r/c$ perpendicular to $\widehat{\mathbf a}$ with $|\mathbf q|\le1$, choose
$$
\mathbf u=\widehat{\mathbf a}\times\mathbf q
+\sqrt{1-|\mathbf q|^2}\,\widehat{\mathbf a}.
$$
The <vector triple product> gives $\mathbf u\times\widehat{\mathbf a}=\mathbf q$, and $|\mathbf u|=1$. Thus every point of the \b[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.
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