Answer
$x-y-z=4$
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:
$-3(x-6)+3(y+1)+3(z-3)=0$
After simplification, we get
$(x-6)-(y+1)-(z-3)=0$
Or, $x-y-z=4$