I am not really able to understand the question. I didn’t get the part ‘contains the projection of the line on the plane....’
How can a plane contain the projection of a line on another plane? Wouldn’t that just mean that required plane is coincident with $2x+3y-z=5$?
I know there is already such a question on MSE but I am not convinced with that answer.
The first vector you provide, $2i+2j+3k$, is the direction of the line $\frac{x-3}{2}=\frac{y+2}{2}=\frac{z-1}{3}$. Surely the plane containing the line should contain this.
The second vector is the "direction" of the plane $2x+3y-z=5$, which is perpendicular to it. And if you can draw a picture on the draft, you know that the plane we want should contain this, since the projected line, the original line, and the perpendicular vector $2i+3j-k$ should be on the same plane.
Finally you need the plane to contain $(3,-2,1)$ which is on the original line. One point and two different vectors is now sufficient to determine a plane--
$a(x-3)+b(y+2)+c(z-1)=0$ is what you want. Here $(a,b,c)\perp(2,2,3)$ and $(a,b,c)\perp(2,3,-1)$ as we discussed above. You can then either solve the equation directly or use outer product.