= Solution
Let $d=b\cdot x$. Taking a <dot product> with $b$ reduces the <vector> equation to $(1-b\cdot c)d=b\cdot a$. If $b\cdot c\ne1$, this determines a unique <scalar> and hence a unique <vector>:
$$
\boxed{x=a+\frac{b\cdot a}{1-b\cdot c}\,c.}
$$
Direct substitution verifies it, so necessity has also produced sufficiency. The formula includes $c=0$, when $x=a$.
If $b\cdot c=1$, the reduced <scalar> equation becomes $0=b\cdot a$. When $b\cdot a\ne0$, \b[there is no solution]. When $b\cdot a=0$, \b[all solutions are $x=a+dc$ with arbitrary $d\in\mathbb R$]: their <dot product> with $b$ is indeed $d$. Since $b\cdot c=1$ forces $c\ne0$, this is an entire affine <line>, not a single point. These alternatives are the <solvability of an affine rank-one vector equation>.
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