Find the equation of the plane through the points (2, 1, 1) and (1, 3, 4) and perpendicular to the plane `x\+ \ 2y+4z=10.`

Explanation

The equation of the plane through `(2,1,-1)` is
`a(x-2)+b(y-1)+c(z+1)=0`
Since this passes through `(-1,3,4)`
`:. a(-1-2)+b(3-1)+c(4+1)=0`
`implies-3a+2b+5c=0`
Since the plane (i) is perpendicular to the plane `x-2y+4z=10`
`:. 1.a-2.b+4.c=0`
`implies a-2b+4c=0`
On solving eqs (ii) and (iii) we get
`a/(8+10)=(-b)/(-17)=c/4=lamda`
`implies a=18lamda,b=17lamda,c=4lamda`.
From eq. (i)
`18lamda(x-2)+17lamda(y-1)+4lamda(z+1)=0`
`implies18x-36+17y-17+4z+4=0`
`implies 18x+17y+4z-49=0`
`. 18x+17y+4z=49`