Find the equation of the plane passing through the points `(1,0,-1)a n d(3,2,2)` and parallel to the line `x-1=(1-y)/2=(z-2)/3dot`

Explanation

The equation of any plane through the point `(1, 0, -1)` is
`" "A(x-1)+B(y-0)+C(z+1)=0" "`(i)
Since it passes through the point `(3, 2, 2)`, we get
`" "2A+2B+3C=0" "`(ii)
Since plane (i) is parallel to the line
`" "(x-1)/(1)=(y-1)/(-2)=(z-2)/(3)`
we have `1A+(-2)B+3C=0" "` (iii)
From (i) and (iii),
`" "A:B:C= 4:-1 : -2`
Substituting these values in (i), we get
`" "4(x-1)-1(y-0)-2(z+1)=0`
or `" "4x-y-2z-6=0`