Why don't these two normal vectors have a 0 or a 180 degree angle between them when graphed?

Question

There's a plane that is given with the equation $2x-3y+z+2=0$ and there's a line equation $\frac{x+3}{-1}=\frac{y}{3}=\frac{z-1}{P}$. I need to find the missing number $P$, when the line is parallel to the plane. As number $P$ is a coordinate of the normal of the line and we also have the normal of the plane, I assumed we could use the dot product here and write $n_1 \cdot n_2 = 0$ as the normals need to be parallel in order for that to happen. I solved that and I got $P=11$, but when I graph the normals they don't seem to be parallel at all. Is there something wrong? Please help me. Thanks in advance.

Answer 1

$v=(-1,3,P)$ is the direction vector of the line and not the normal (which would be a plane).

And indeed, the solution is to find $P$ for which $n\cdot v=0$ and $P=11$ seems correct.

To verify, show that the line and the plane don't have any common points.