Question

Consider the planes
$P_1:cy+bz=x$
$P_2:az+cx=y$
$P_3:bx+ay=z.$
$P_1,P_2$ and $P_3$ pass through one line, if

• A. $a^2+b^2+c^2=ab+bc+ca$
• B. $a^2+b^2+c^2+2abc=1$
• C. $a^2+b^2+c^2=1$
• D. $a^2+b^2+c^2+2ab+2bc+2ca+2abc=1$

Answer 1

The correct option is B $a^2+b^2+c^2+2abc=1$

Given planes
$P_1:cy+bz-x=0$
$P_2:az+cx-y=0$
$P_3:bx+ay-z=0$

Above planes can be written as
$P_1:-x+cy+bz=0...\left(1\right)$
$P_2:cx-y+az=0...\left(2\right)$
$P_3:bx+ay-z=0...\left(3\right)$
This is a homogeneous system of equations.
As all planes pass through one line, it means this system has infinitely many solutions.
So, $\left|\begin{array}{ccc}-1& c& b\\ c& -1& a\\ b& a& -1\end{array}\right|=0$

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