Question

The planes $x=cy+bz,y=az+cx,z=bx+ay$ pass through one line, if

• A. $a+b+c=0$
• B. $a+b+c=1$
• C. $a^2+b^2+c^2=1$
• D. $a^2+b^2+c^2+2abc=1$

Answer 1

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

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

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

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

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

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

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

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

$⟹\frac{y}{x}=\frac{(c+ab)}{(1-a^2)}-----\left(5\right)$

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

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

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