 Solution

Apply the vector triple product identity and the two plane constraints:Put $d=b\times c\ne 0$. 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 thereforeSuch 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.