 Solution

Substitute $\mathbf{x}=\mathbf{a}+\lambda \mathbf{c}$ into the original equation and cancel $\mathbf{a}$. Because $\mathbf{c}$ is nonzero, the scalar condition isIf $\mathbf{b}\cdot \mathbf{c}\ne 1$, there is one value, $\lambda =(\mathbf{b}\cdot \mathbf{a})/(1-\mathbf{b}\cdot \mathbf{c})$. If $\mathbf{b}\cdot \mathbf{c}=1$ and $\mathbf{b}\cdot \mathbf{a}=0$, every real $\lambda$ works. If $\mathbf{b}\cdot \mathbf{c}=1$ and $\mathbf{b}\cdot \mathbf{a}\ne 0$, the scalar equation is inconsistent and there is no solution.

Thus “undefined” in the last case means that no parameter value solves the equation; it must not be used to conceal the distinct case with infinitely many solutions.