Normals are an equation sorta - they're the vector equation that points out perpendicular to the surface of the plane. I assume they're giving you the x,y,z components in the 'sets of data'.
Here's basically what you're gonna want to do: take your line and parameterize it. So if your line goes from (1,2,3) to (8,4,9), the equation for your line is (1+7t,2+2t,3+6t), where t is 0 to 1. This will make it significantly easier to solve the rest of the problem. This might have already been done depending on what 'equation for the line' you have.
Then take your normals and the 3d points they come with and solve for the equation of your plane. Basically the normals tell you everything about the slope of your plane in three different axes, but the 3d points tell you where to actually put that plane. If your normal vectors are given as (Nx, Ny, Nz) and your 3d points are (Px, Py, Pz), the formula for your plane is going to be:
(X - Px)Nx + (Y-Py)Ny + (Z-Pz)Nz = 0
Rearrange it so that all the constants are on the right side.
Then simply take whatever plane equation you have, maybe something like X + Y + Z = 3, and substitute in your line's parameters, which might be something like X = 2t, Y = 2 + 3t, Z = 4t, and then solve for t. So in this case, that's:
(2t) + (2+3t) + (4t) = 3
9t + 2 = 3
9t = 1
t = 1/9, and you plug that back into your parametric equation to get (2/9,2+3/9,4/9), which is the point where that line intersects the plane.