Question

Given 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 $a^2+b^2+c^2+2abc=1$

Since,$x=cy+bz---\left(1\right)$

$y=az+cx----\left(2\right)$

$z=bx+ay----\left(3\right)$

by cancellation z from (1)&(2) by help (3) we get,

$(1-b^2)x=(c+ab)y$

$\frac{x}{y}=\frac{(c+ab)}{(1-b^2)}$----(4)

$\frac{y}{x}=\frac{c+ab}{1-a^2}$

now by multiplying (4) and (5) we get

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

$\therefore a^2+b^2+c^2+2abc=1.$