= 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 is
$$
\boxed{\lambda(1-\mathbf b\cdot\mathbf c)=\mathbf b\cdot\mathbf a}.
$$
If $\mathbf b\cdot\mathbf c\ne1$, 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 $\mathbf b\cdot\mathbf a\ne0$, 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.
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