If planes $x-cy-bz=0, cx-y+az=0$ and $bx+ay-z=0$ pass through a straight line then $a^2+b^2+c^2=$

• A

  $1-abc$
• B

  $abc-1$
• Correct answer
  C

  $1-2abc$
• D

  $2abc-1$

Solution

Since the given planes pass through a straight line

$\Rightarrow \left|\begin{array}{ccc}1& -c& -b\\ c& -1& a\\ b& a& -1\end{array}\right|=0$
$\Rightarrow 1\left(1-a^2\right)+c\left(-c-ab\right)-b\left(ac+b\right)=0$
$\Rightarrow 1-a^2-c^2-abc-abc-b^2=0$
$\Rightarrow 1-2abc-a^2-b^2-c^2=0$
$\Rightarrow 1-2abc=a^2+b^2+c^2$
 

Hence, the correct answer is option: 'C'.