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

$x+3y-z=9$
Normal to the plane: $n=\lt -1,-3,1 \gt$ 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 $(2,4,5)$, we have $(-1)(x-2)+(-3)(y-4)+(1)(z-5)=0$ or, $-x-3y+z=-9$ or, $x+3y-z=9$