Find the equation of a plane which passes through the point `(1,2,3)` and which is equally inclined to the planes `x-2y+2z-3=0a n d8x-4y+z-7=0.`

Explanation

Plane which is equally inclined to the given planes is parallel to the angle bisector of the given planes.
Now the angle bisector of the given planes is
`" "(x-2y+2z-3)/(3)=pm (8x-4y+z-7)/(9)`
`" "5x+2y-5z+2=0`
and `" "11x-10y+7z-16=0`.
The equation of the required planes are `5x+2y-5z+p=0 and 11x-10y+7z+q=0`.
Since both are passing through point `(1, 2, 3), p=6 and q=12`.
The planes are `5x+2y-5z+6=0 and 11x-10y+7z+12 =0 `