Intersection of a plane and a normal.

Problem: A normal to $\displaystyle 3x-y+4z=12 $passes through A(5,1,6). Find the point of intersection of the normal and the plane.

I haven't solved a question like this before in my book, so I am not too sure on how to go about this question. I know that the normal vector to $\displaystyle 3x-y+4z=12 $ is $\displaystyle \vec{n} = (3, -1, 4)$. So, to find the point of intersection, do I take the cross product of the normal and the given point? Or, would I have to find another point on the plane (I'll let $\displaystyle x = 2$ and $\displaystyle y = 1$ to get

$\displaystyle 3(2) - (1)+4z = 12 => 4z = 7 => z = \frac{7}{4}$, this point will be $\displaystyle B(2, 1,\frac{7}{4})$ and then find $\displaystyle \vec{AB}$ and take the cross product of $\displaystyle \vec{n}$ and $\displaystyle \vec{AB}$?

I just never encountered a question like this before, so I would appreciate some help on getting started. Thanks in advance.