A) \[-1\]
B) \[2\]
C) \[1\]
D) \[0\]
Correct Answer: B
Solution :
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\[\lambda =-\frac{(ab+c)}{(1-{{a}^{2}})}\] \[\therefore \] \[-\frac{(a+bc)}{(ac+b)}=-\frac{(ab+c)}{(1-{{a}^{2}})}\] \[\Rightarrow \]\[a-{{a}^{3}}+bc-{{a}^{2}}bc={{a}^{2}}bc+a{{c}^{2}}+a{{b}^{2}}+bc\] \[\Rightarrow \]\[2{{a}^{2}}bc+a{{c}^{2}}+a{{b}^{2}}+{{a}^{3}}-a=0\] \[\Rightarrow \]\[a(2abc+{{c}^{2}}+{{b}^{2}}+{{a}^{2}}-1)=0\] \[\Rightarrow \]\[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2abc=1\]
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