If the planes $x=cy+bz,\:\:y=az+cx,\:z=bx+ay$ pass through a line then $a^2+b^2+c^2+2abc=?$

Question

$\begin{array}{1 1} (a)\:\:0\:\:\:\:\qquad\:\:(b)\:\:1\:\:\:\:\qquad\:\:(c)\:\:2\:\:\:\:\qquad\:\:(d)\:\:-1. \end{array}$</p

Comments

Ans is 1.
Since, x=cy+bz----(1)
y=az+cx----(2)
and z=bx+ay----(3)
by cancellation z from (1)&(2) by help (3) we get,
(1-b^2)x=(c+ab) y
=>x/y=(c+ab)/(1-b^2) ----(4)
&(1-a^2)y=(c+ab) x
=>y/x=(c+ab)/(1-a^2) -----(5)
now by multiplying(4)&(5) we get,
1=(c^2+2abc+a^2b^2)/(1-a^2-b^2+a^2b^2)
=> a^2+b^2+c^2+2abc=1.