Question

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

• A. $-1$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is C $1$

Given planes are
$x-cy-bz=0$ ..... (i)
$cx-y+az=0$ .... (ii)
$bx+ay-z=0$ ..... (iii)
Equation of planes passing through the line of intersection of planes (i) and (ii) may be taken as
$(x-cy-bz)+\lambda (cx-y+az)=0$
or $x(1+\lambda c)-y(c+\lambda )+z(-b+a\lambda )=0$ .... (iv)
If planes (iii) and (iv) are same, then Eqs. (iii) and (iv) will be identical.
$\frac{1+x\lambda }{b}=\frac{-(c+\lambda )}{a}=\frac{-b+a\lambda }{-1}$
$\Rightarrow \lambda =-\frac{(a+bc)}{(ac+b)}$ and $\lambda =-\frac{(ab+c)}{(1-a^1)}$
Therefore, $-\frac{(a+bc)}{(ac+b)}=-\frac{(ab+c)}{(1-a^2)}$
$\Rightarrow a-a^3+bc-a^{2}bc=a^{2}bc+a{c}^2+a{b}^2+bc$
$\Rightarrow 2a^{2}bc+a{c}^2+a{b}^2+a^3-a=0$
$\Rightarrow a(2abc+c^2+b^2+a^2-1)=0$
$\Rightarrow a^2+b^2+c^2+2abc=1$.