Question

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

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

$P_1:x=cy+bz$,

$P_2:y=az+cx$ and

$P_3:z=bx+ay$

If they pass through a line
$P_3=P_1+\lambda P_2$
$z-bx+ay=x-cy+bz+\lambda (y-az+cx)$

Comparing the coefficients of $x,y,z$ we get
$1=b-\lambda a$
$-b=1+\lambda c$
$a=-c+\lambda$
Solving, we get,
$b=1$ and
$a+c=0$
So

$a^2+b^2+c^2+2abc=(a+c{)}^2+1=1$

Hence, option C.