= 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, $\mathbf c_1\times\mathbf c_3=(-7,1,3)$, so
$$
[\mathbf c_1,\mathbf c_3,\mathbf b]=(-7,1,3)\cdot(1,\lambda,0)=\lambda-7.
$$
Thus $\lambda=7$ is necessary and, since the two columns span the plane, sufficient. The equations then give $x-y=1$ and $z=2$, so
$$
\boxed{\lambda=7,\qquad\mathbf r=(1,0,2)^{\mathsf T}+t(1,1,0)^{\mathsf T},\quad t\in\mathbb R.}
$$
The 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 $z=2$.
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