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} $.