The question asks me to find the inverse map $ \mathbf\Phi^{-1} $, of:
$$ \mathbf{\Phi}(\mathbf{x})= \mathbf{n\lor(x \lor n)} + \alpha\mathbf{(n \cdotp x)n} $$ for $\alpha$ such that the inverse exists. I'm not quite sure how to do this without finding the matrix representing the map and then taking its inverse (for $\alpha$ such that it exists) which in this case would be rather messy. I've tried setting $ \mathbf{x'} = \mathbf{\Phi}(\mathbf{x})$ and rearranging but this only yields: $$ \mathbf{x'}= \mathbf{x} -(1-\alpha)(\mathbf{x \cdot n})\mathbf{n} $$ and I don't see how this will let me express $\mathbf{x}$ explicitly in terms of $\mathbf{x'}$ to find the inverse function, without dotting or crossing either side, which I believe is not allowed as such a step isn't reversible.
I assume that $\mathbf{n}$ is a unit vector.
It is easy to see that, when $\alpha=1$, the two terms of the sum are the components of the vector $\mathbf{x}$ parallel and orthogonal to $\mathbf{n}$. So in such a case their sum is exactly $\mathbf{x}$, i.e. $\Phi(\mathbf{x})=\mathbf{x}$.
When $\alpha\neq 1$, the component parallel to $\mathbf{n}$ is incremented by a factor $\alpha$, so the inverse is obtained simply decrementing such a component by a factor $1/\alpha$: $$ \mathbf{x} = \mathbf{\Phi}^{-1}(\mathbf{y}) = \mathbf{n} \wedge (\mathbf{y} \wedge \mathbf{n}) + \frac{1}{\alpha}\,(\mathbf{n} \cdot \mathbf{y})\,\mathbf{n} $$
The case to exclude is $\alpha=0.$