= Solution
Apply the <vector triple product identity> and the two <plane> constraints:
$$
\boxed{r\times(b\times c)=b(r\cdot c)-c(r\cdot b)=\mu b-\lambda c.}
$$
Put $d=b\times c\ne0$. If $a$ is one intersection point, subtraction gives $(r-a)\times d=0$, so $r-a$ is parallel to $d$. Conversely, adding any multiple of $d$ preserves both <plane> constraints because $d$ is <orthogonal> to both <normal vectors>. The whole intersection is therefore
$$
\boxed{r=a+t(b\times c),\quad t\in\mathbb R,\qquad a\cdot b=\lambda,\quad a\cdot c=\mu.}
$$
Such an $a$ exists: choose $a=\alpha b+\beta c$ and solve the two-by-two <linear system> with <determinant> $|b|^2|c|^2-(b\cdot c)^2=|b\times c|^2>0$. This also proves sufficiency rather than assuming a nonempty intersection.
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