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

Answer: (B)

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 $\lambda =-\frac{(ab+c)}{(1-a^2)}$
$\therefore$ $-\frac{(a+bc)}{(ac+b)}=-\frac{(ab+c)}{(1-a^2)}$
$\Rightarrow$ $a-a^3+bc-a^{2}bc=a^{2}bc+a{c}^2+a{b}^2+bc$
$\Rightarrow$ $2a^{2}bc+a{c}^2+a{b}^2+a^3-a=0$
$\Rightarrow$ $a(2abc+c^2+b^2+a^2-1)=0$
$\Rightarrow$ $a^2+b^2+c^2+2abc=1$