Answer
$x+3y+2z= 3$
Work Step by Step
Scalar form of the equation of a plane is
$a(x-x_{0})+b(y-y_{0})+c(z-z_{0})=0$
where $(x_{0},y_{0},z_{0})$ is the point; a,b and c are direction ratios of the normal vector to the plane.
Substituting the values, we have
$1(x-4)+3(y-(-1))+2(z-1)=0$
or $x-4+3y+3+2z-2=0$
or $x+3y+2z=3$
