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 $ $ 2{{a}^{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 $