Answer
$x-2y+3z=-13$
Work Step by Step
The standard equation of a plane passing through the point $(x_1,y_1,z_1)$ can be defined as:
$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$
The normal vector to the plane is given by:$n=\lt 1,-2,3 \gt$
Thus, $1(x+1)-2(y -6)+3(z-0)=0 \implies x+1-2y+12+3z=0$
or, $x-2y+3z=-13$
