 Solution

At $\mu =-1$, ${\mathbf{c}}_2=-{\mathbf{c}}_1$, while ${\mathbf{c}}_1$ and ${\mathbf{c}}_3$ are independent. Their span is a plane. A solution requires $\mathbf{b}$ to lie in that plane, so every scalar triple product formed from two columns and $\mathbf{b}$ must vanish. The product with ${\mathbf{c}}_1,{\mathbf{c}}_2$ vanishes identically; the other two give the plane condition.

Explicitly, ${\mathbf{c}}_1\times {\mathbf{c}}_3=(-7,1,3)$, soThus $\lambda =7$ is necessary and, since the two columns span the plane, sufficient. The equations then give $x-y=1$ and $z=2$, soThe solution is an affine line parallel to the homogeneous null space from part (c), obtained by intersecting the compatible constraint planes. Every point on it is nonzero because $z=2$.