Same thing, really: a vector <w, x , y, z> is perpendicular to <3, -5, 7, 9> if and only if there dot product, 3w- 5x+ 7y+ 9z= 0.
You can solve for any one of those variables in terms of the other 3, then write <w, x, y, z> in terms of those three.
Rather than do that problem for you consider the space of all vector perpendicular to <-2, 4, 6, 5>. That would consist of all vectors <w, x, y, z> such that -2w+ 4x+ 6y+ 5z= 0. Solving for w, w= 2x+ 3y+ (5/2)z. Then, <w, x, y, z>= <2x+ 3y+ (5/2)z, x, y, z>= <2x, x, 0, 0> + <3y, 0, y, 0> + <(5/2)z, 0, 0, z>= x<2, 1, 0, 0>+ y<3, 0, 1, 0> + z<5/2, 0, 0, 1>.
And now, a basis is obvious.
If you don't like fractions, that last one can be written (z/2)<5, 0, 0, 2>