## University Calculus: Early Transcendentals (3rd Edition)

$x-y+z=0$
The normal to the plane is $n=\lt 3,-3,3 \gt$ We know that the standard equation of a plane passing through the point $(x_0,y_0,z_0)$ is written as: $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ Then for point $(2,1,-1)$, we have $3(x-2)-3(y-1)+3(z+1)=0$ or, $3x-3y+3z=0$ or, $x-y+z=0$