= Solution
The <homogeneous linear system> has coefficient <matrix>
$$
M=\begin{pmatrix}1&0&-3\\1&\lambda&-\lambda\\1&2&\lambda\end{pmatrix}.
$$
Nonzero solutions exist exactly when its <determinant> vanishes. Expanding along the first row gives
$$
\det M=\lambda(\lambda+2)-3(2-\lambda)=\lambda^2+5\lambda-6=(\lambda-1)(\lambda+6).
$$
Thus $\lambda=1$ or $-6$. The first equation always gives $x=3z$. For $\lambda=1$, the second then gives $y=-2z$; the third imposes the same relation. For $\lambda=-6$, the second gives $y=3z/2$, again consistent with the third. Therefore the nonzero solutions are
$$
\boxed{\begin{cases}\lambda=1:&(x,y,z)=t(3,-2,1),\\\lambda=-6:&(x,y,z)=t(6,3,2),\end{cases}\qquad t\in\mathbb R\setminus\{0\}.}
$$
For every other $\lambda$, the <matrix> is invertible and only the zero solution exists.
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