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 a2+b2+c2+2abc is
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)+λ(cx−y+az)=0i.e. x(1+λc)−y(c+λ)+z(−b+aλ)=0...........(iv)
If plane (iii) and (iv) are the same, then equation (iii) and (iv) will be identical.
⇒1+xλb=−(c+λ)a=−b+aλ−1⇒λ=−(a+bc)ac+b and λ=−(ab+c)1−a2∴−(a+bc)(ac+b)=−(ab+c)(1−a2)⇒a−a3+bc−a2bc=a2bc+ac2+ab2+bc⇒2a2bc+ac2+ab2+a3−a=0⇒a(2abc+c2+b2+a2−1)=0⇒a2+b2+c2+2abc=1 (∵ a≠0)