Answer
$x-2y+3z=-13$
Work Step by Step
The normal to the plane is $n=\lt 1,-2,3 \gt$
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, we have the point $P(-1,6,0)$
$1(x+1)-2(y -6)+3(z-0)=0$
or, $x+1-2y+12+3z=0$
or, $x-2y+3z=-13$
