Answer
$2x+y+3z=3$
Work Step by Step
The general form of the equation of the plane passing through the point $(a,b,c)$ and having normal vector $\lt l,m,n\gt$ is:
$l(x-a)+m(y-b)+n(z-c)=0$
Thus, the equation of the plane is:
$2(x-3)+1(y-0)+3(z+1)=0$
After simplification, we get
$2x+y+3z-3=0$
Or, $2x+y+3z=3$
