Question

The planes x-cy-bz=0, cx-y+az=0 and bx+ay-z=0 pass through a straight line, where a,b,c are non-zero constants. 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 pf planes passing through the line of intersection of plane (i) and (ii) may be taken as;
$(x-cy-bz)+\lambda (cx-y+az)=0i.e.x(1+\lambda c)-y(c+\lambda )+z(-b+a\lambda )=0...........\left(iv\right)$
If plane (iii) and (iv) are the same, then equation (iii) and (iv) will be identical.
$\Rightarrow \frac{1+x\lambda }{b}=\frac{-(c+\lambda )}{a}=\frac{-b+a\lambda }{-1}\Rightarrow \lambda =-\frac{(a+bc)}{ac+b}and\lambda =-\frac{(ab+c)}{1-a^2}\therefore \frac{-(a+bc)}{(ac+b)}=-\frac{(ab+c)}{(1-a^2)}\Rightarrow a-a^3+bc-a^{2}bc=a^{2}bc+a{c}^2+a{b}^2+bc\Rightarrow 2a^{2}bc+a{c}^2+a{b}^2+a^3-a=0\Rightarrow a(2abc+c^2+b^2+a^2-1)=0\Rightarrow a^2+b^2+c^2+2abc=1(\because a\ne 0)$