Solution
= Solution
For $\lambda\ne\pm2$, the reduced equation gives $y=(3-\lambda)/(4-\lambda^2)$. Back substitution into $x=1-\lambda y$ and $z=1-4y$ gives all solutions:
$$
\boxed{\boldsymbol r=
\frac1{4-\lambda^2}\begin{pmatrix}
4-3\lambda\\3-\lambda\\-\lambda^2+4\lambda-8
\end{pmatrix},\qquad \lambda\ne\pm2.}
$$
The nonzero <determinant> proves that there are no other solutions at these parameter values.