The planes defined by
\begin{align*}
-x + cy + bz &= 0, \\
cx - y + az &= 0, \\
bx + ay - z &= 0
\end{align*}all contain the same line.  Find all possible values of
\[a^2 + b^2 + c^2 + 2abc.\]Enter all the possible values, separated by commas.
The normal vectors of the planes are $\mathbf{n}_1 = \begin{pmatrix} -1 \\ c \\ b \end{pmatrix},$ $\mathbf{n}_2 = \begin{pmatrix} c \\ -1 \\ a \end{pmatrix},$ and $\mathbf{n}_3 = \begin{pmatrix} b \\ a \\ -1 \end{pmatrix}.$  So, the direction vector of the common line is proportional to
\[\mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix} ac + b \\ a + bc \\ 1 - c^2 \end{pmatrix}.\]It is also proportional to
\[\mathbf{n}_1 \times \mathbf{n}_3 = \begin{pmatrix} -ab - c \\ b^2 - 1 \\ -a - bc \end{pmatrix}.\]Since these vectors are proportional,
\[(ac + b)(b^2 - 1) = (a + bc)(-ab - c).\]Then $(ac + b)(b^2 - 1) - (a + bc)(-ab - c) = 0,$ which simplifies to
\[a^2 b + 2ab^2 c + b^3 + bc^2 - b = 0.\]This factors as $b(a^2 + b^2 + c^2 + 2abc - 1) = 0.$

Similarly,
\[(ac + b)(-a - bc) = (1 - c^2)(-ab - c).\]This becomes $c(a^2 + b^2 + c^2 + 2abc - 1) = 0.$

If both $b = 0$ and $c = 0,$ then the equations of the planes become
\begin{align*}
x &= 0, \\
-y + az &= 0, \\
ay - z &= 0.
\end{align*}Then $y = az.$  Substituting into the third equation, we get $a^2 z - z = 0,$ so $(a^2 - 1) z = 0.$  If $a^2 \neq 1,$ then we must have $z = 0,$ which leads to $y = 0,$ so the three planes only have the point $(0,0,0)$ in common.  Hence, $a^2 = 1.$  Then the equations of the planes become $x = 0,$ $y = z,$ and $y = z,$ and their intersection is a line.  Also,
\[a^2 + b^2 + c^2 + 2abc = 1.\]Otherwise, at least one of $b$ and $c$ is nonzero, so $a^2 + b^2 + c^2 + 2abc - 1 = 0.$  Hence,
\[a^2 + b^2 + c^2 + 2abc = 1.\]We conclude that $a^2 + b^2 + c^2 + 2abc$ is always equal to $\boxed{1}.$