Finding the equation of a plane

Hey, I was hoping you kind people here at mathhelpforum.com could help me out with a problem that I am stuck on.

** Question **

Find the equation of the plane that passes through the line of intersection of the planes 3x - y + z - 2 = 0 and x + 2y - 4z + 1 = 0, and that satisfies the given condition.

a) the plane passes through the point A(3,1,3)

b) the plane is parallel to the plane 5x + 3y - 7z - 6 = 0

** Attempt at a solution **

So taking the cross product of the 2 given planes gives the direction vector (2,13,7). Then setting a variable to 0, I chose x, I get the vector equation: $\displaystyle r = (0,\frac{-7}{2},\frac{3}{-2}) + t(2,13,7) $.

a)So, knowing that, how do I find the plane that goes through (3,1,3)? If I had the normal to the plane I could do it, so how do I find the normal to (2,13,7)?

b) The dot product of (5,3,-7) and (2,13,7) = 0. Which means I can sub in a point from the line of intersection in for x y and z, and solve for D. A point from the line of intersection is $\displaystyle (0,\frac{-7}{2},\frac{3}{-2})$. Solving for D gives me 21. So the equation of the plane that passes through $\displaystyle r = (0,\frac{-7}{2},\frac{3}{-2}) + t(2,13,7) $ and is parallel to [tex] 5x + 3y - 7z - 6 = 0 [tex] is $\displaystyle 5x + 3y - 7z + 21 $.

I am unsure if either of those are the correct way of doing it. If not, can someone put me on the right track?

Thank you so much for reading everyone! Hopefully you can help me out.