Answer
$x+3y-z=9$
Work Step by Step
The standard equation of plane passing through the point $(a,b,c)$ is given as $p(x-a)+q(y-b)+r(z-c)=0$
Here, the equation of normal to the plane is $n=\lt -1,-3,1 \gt$
For point $(2,4,5)$, we have
$(-1)(x-2)+(-3)(y-4)+(1)(z-5)=0$
or, $-x-3y+z=-9$
Thus, $x+3y-z=9$
