Question

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

• A. $-1$
• B. $0$
• C. $1$
• D. None of these

Answer 1

The correct option is C $1$

Let $l,m,n$ be the direction ratios of the line lying in the three planes, then the line is perpendicular to the normals to these planes, so that we have

$l-cm-bn=0$

$-cl+m-an=0$

$-bl-am+n=0$

Eliminating $l,m,n$ from these equations, we get

$\left|\begin{array}{ccc}1& -c& -b\\ -c& 1& -a\\ -b& -a& 1\end{array}\right|=0$

$\Rightarrow 1-a^2+c(-c-ab)-b(ac+b)=0\Rightarrow a^2+b^2+c^2+2abc=1$