= Solution
The equations are an <eigenvalue problem> for the <matrix>
$$
A=\begin{pmatrix}2&-2&1\\2&-3&2\\-1&2&0\end{pmatrix}.
$$
A nonzero solution requires $\det(\lambda I-A)=0$. Expanding gives
$$
\det(\lambda I-A)=(\lambda-1)^2(\lambda+3),\qquad \boxed{\lambda=1\ \text{or}\ -3.}
$$
<Row reduction> gives the complete <eigenspaces>:
$$
\boxed{\begin{aligned}
\lambda=1:&\quad (x,y,z)=s(2,1,0)+t(-1,0,1),\qquad s,t\in\mathbb R,\\
\lambda=-3:&\quad (x,y,z)=u(-1,-2,1),\qquad u\in\mathbb R.
\end{aligned}}
$$
For $\lambda=1$ the solutions form the <plane> $x-2y+z=0$ through the origin; for $\lambda=-3$ they form a <line> through the origin. Exclude the zero vector when requiring a nontrivial solution. The repeated <eigenvalue> $1$ has a two-dimensional <eigenspace>; there are three independent <eigenvectors> in total. This matrix is not symmetric, so its different <eigenspaces> need not be mutually <orthogonal>.
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