= Solution
Collecting the coefficients gives the <coefficient matrix>
$$
\boxed{A=\begin{pmatrix}a&0&2\\0&b+1&c-1\\0&c+1&b-1\end{pmatrix},\qquad
A\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}.}
$$
A square <homogeneous linear system> has a nonzero solution precisely when its <matrix> is a <singular matrix>. Expanding the <determinant> down the first column,
$$
\det A=a[(b+1)(b-1)-(c-1)(c+1)]=a(b^2-c^2).
$$
Hence the necessary and sufficient condition is \b[$a=0$ or $b=c$ or $b=-c$]. A zero <determinant> gives a nontrivial <kernel of a linear map>; otherwise multiplication by the <matrix inverse> would force the zero solution.
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