Question

If the planes $x=cy+bz,y=az+cx$ and $z=bx+ay$ pass through a line, then $a^2+b^2+c^2+2abc+4=$

Answer 1

Let $l,m,n$ be the d.c.'s of the line through which the given planes pass. This line will be
perpendicular to normals of the given planes
$-l+cm+bn=0$
$cl-m+an=0$
$bl+am-n=0$
Eliminating $l,m,n$, we have
$\left|\begin{array}{ccc}-1& c& b\\ c& -1& a\\ b& a& -1\end{array}\right|=0$
$-1(1-a^2)-c(-c-ba)+b(ac+b)=0$
$a^2+b^2+c^2+2abc-1=0$

$a^2+b^2+c^2+2abc+4=5$