Question

If the planes $x-cy-bz=0,cx-y+az=0$ and $bx+ay-z=0$ pass through a stright line,then the value of $a^2+b^2+c^2+2abc$ is:

• A. $1$
• B. $2$
• C. $3$
• D. none of these

Answer 1

The correct option is A $1$

If the planes $(x-cy-bz=0),(cx-y+az=0)$ and $(bx+ay-z=0)$ are in same line.

$\therefore$ They must be collinear.

$\left|\begin{array}{ccc}1& -c& -b\\ c& -1& a\\ b& a& -1\end{array}\right|=0$

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

$\Rightarrow 1-a^2-c^2-abc-abc-b^2=0$

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

Option A is correct