Find the equations of the bisectors of the angles between the planes `2x-y+2z+3=0` and ` 3x-2y+6z+8=0` and specify the plane which bisects the acute angle and the plane which bisects the obtuse angle.

Explanation

The given planes are `2x-y+2z+3=0 and 3x-2y+6z+8=0` , where `d_(1), d_(2) gt 0 and a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)= 6+2+12gt 0`.
`" "(a_(1)x+b_(1)y+c_(1)z+d_(1))/(sqrt(a_(1)^(2)+b_(1)^(2)+c_(1)^(2)))=-(a_(2)x+b_(2)y+c_(2)z+d_(2))/(sqrt(a_(2)^(2)+b_(2)^(2)+c_(2)^(2)))" "` (obtuse angle bisector)
`" "(a_(1)x+b_(1)y+c_(1)z+d)/(sqrt(a_(2)^(2)+b_(2)^(2)+c_(2)^(2)))=(a_(2)x+b_(2)y+c_(2)z+d_(2))/(sqrt(a_(2)^(2)+b_(2)^(2)+c_(2)^(2)))" "` (acute angle bisector)
i.e., `" "(2x-y+2z+3)/(sqrt(4+1+4))=pm(3x-2y+6z+8)/(sqrt(9+4+36))`
or `" "(14x-7y+14z+21)=pm(9x-6y+18z+24)`
Taking the positive sign on the right hand side, we get
`" "5x-y-4z-3=0`
Taking the negative sign on the right hand side, we get
`" "23x-13y+32z+45=0" "` (acute angle bisector)