If the planes `x-c y-b z=0,c x-y+a z=0a n d b x+a y-z=0` pass through a straight line, then find the value of `a^2+b^2+c^2+2a b c dot`

Answer

1

Explanation

The given planes are
`x-cy-bz=0" "(i)`
`cx-y+az=0" "(ii)`
`bx+ay-z=0" "(iii)`
The equation of the planes passing through the line of intersection of planes (i) and (ii) may be taken as
`(x-cy-bz)+lamda(cx-y+az)=0`
or `x(1+lamdac)-y(c+lamda)+z(-b+alamda)=0" "(iv)`
If planes (iii) and (iv) are the same, then Eps. (iii) and (iv) will be identical.
`(1+clamda)/(b)=(-c+lamda)/(a)=(-b+alamda)/(-1)`
`lamda=-(a+bc)/(ac+b)andlamda=-(ab+c)/(1-a^(2))`
`because(-a+bc)/(ac+b)=-(ab+c)/(1-a^(2))`
`a-a^(3)+bc-a^(2)bc=a^(2)bc+ac^(2)+ab^(2)+bc`
or `2a^(2)bc+ac^(2)+ab^(2)+a^(3)-a=0`
or `a(2abc+c^(2)+b^(2)+a^(2)-1)=0`
or `a^(2)+b^(2)+c^(2)+2abc=1`
Alternate method:
Since the planes pass through the origin, the given planes have a common line of intersection if given system of equations has a non-trivial solution. Hence,
`|[1,-c,-b],[c,-1,a],[b,a,-1]|=0`
or `a^(2)+b^(2)+c^(2)+2abc=1`