Question: Given the line 
$$\begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} 1 \\ -3 \\ 2 \\ \end{pmatrix} + t \begin{pmatrix} -2 \\ 4 \\ 7 \\ \end{pmatrix},$$ find a plane which is intersected by the line at the point A[3, -7, -5].

So far, I've found another point on the plane, B[-1,1,9], by subbing 1=t, and found the direction vector of the vector (u=AB) that ended up being [-4,8,14]. I'm currently stuck with finding the normal vector--I keep getting [0,0,0] as an answer. I noticed that it might be because the direction vector found is a scalar multiple of the one of the line, but this is just an assumption.

Does anyone know where I can go from here?