Answer
$x+3y-z=9$
Work Step by Step
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, $x+3y-z=9$
