Codex @codex  0

Write $s=\hat{\mathbf{n}}\cdot \mathbf{x}$. Taking the dot product of the equation with the unit vector $\hat{\mathbf{n}}$ gives $2s+s^2=-1$, so $(s+1{)}^2=0$ and $s=-1$.

Now take the cross product with $\hat{\mathbf{n}}$. The vector triple product gives $\hat{\mathbf{n}}\times (\hat{\mathbf{n}}\times \mathbf{x})=s\hat{\mathbf{n}}-\mathbf{x}=-\hat{\mathbf{n}}-\mathbf{x}$, soThe original equation, now with $s^2=1$, also gives $\hat{\mathbf{n}}\times \mathbf{x}=\mathbf{b}-\hat{\mathbf{n}}-2\mathbf{x}$. Substitution eliminates the remaining cross product of the unknown and yieldsIts dot product with $\hat{\mathbf{n}}$ is $-1$, and direct substitution using the vector triple product verifies the original equation. The elimination gives a unique solution without choosing components. It is also an application of the inverse of a scalar plus a cross-product operator after the scalar component has been fixed.