 Closest point on the intersection of two planes

For two planes $r\cdot b=\lambda$ and $r\cdot c=\mu$ with linearly independent normal vectors, put $B=b\cdot b$, $C=c\cdot c$, $D=b\cdot c$ and $\Delta =BC-D^2>0$. Their intersection is a line parallel to $b\times c$. Its closest point to the origin is orthogonal to that direction, hence has the form $a=\alpha b+\beta c$. The constraints give $B\alpha +D\beta =\lambda$, $D\alpha +C\beta =\mu$, soFor every other point $a+t(b\times c)$, orthogonality gives squared distance $|a{|}^2+t^2|b\times c{|}^2$, proving the global minimum and its uniqueness. The positive determinant $\Delta =|b\times c{|}^2$ also proves that an intersection exists for every real $\lambda ,\mu$. Parallel normals require a separate consistency check and do not define a unique intersection line.