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

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

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.

Given equations of the planes are $x-cy-bz=0......(i)$
$cx-y+az=0..............(ii),\:\:\:\:and\:\:\:\:bx+ay-z=0...(iii)$
If a plane passes through a line then it is $\perp$ to the normal of the plane.
Let the $d.r.$ of the line be $(l,m,n)$
Then, $(l,m,n).(1,-c,-b)=0,\:\:(l,m,n).(c,-1,a)=0\:\:\:and\:\:\:(l,m,n).(b,a,-1)=0$
$\Rightarrow\:l-mc-nb=0...(A),\:\:lc-m+na=0...........(B)$ and
$bl+am-n=0.......(C)$
Solving $(A)\:\:\:and\:\:\:(B)$ by cross product method,
$\large\frac{l}{-ac-b}=\frac{-m}{a+bc}=\frac{n}{-1+c^2}$
$\Rightarrow\:l=-ac-b,\:\:m=-a-bc,\:\:n=c^2-1$
Substituting the values in $(C)$ we get $b(-ac-b)+a(-a-bc)-c^2+1=0$
$\Rightarrow\:-2abc-a^2-b^2-c^2+1=0$ or $a^2+b^2+c^2+2abc=1$