Equation of the plane passing through point (1,2,3) is
a(x-1)+ b(y-2)+c(z-3)=0 ..........(1)
Equation of the plane passes through (0,-1,0)
Hence putting in equation in (1)
-a-3b-3c=0.............(2)
The equation of the plane parrallal to the given line
(x-1)/2=(y+2)/3=(z-0)/-3
Normal to the plane is parpendicular to the given line
So 2a+3b-3c=0.........(3)
a=6k, b=-3k, c=k
Putting the value in equation (1)
6k(x-1)+-3k(y-2)+k(z-3)=0
=> 6x-6+3y-6+z-3=0
=> 6x+3y+z=9