In Cartesian space, three spheres centered at $(-2,5,4),$ $(2,1,4),$ and $(4,7,5)$ are all tangent to the $xy$-plane.  The $xy$-plane is one of two planes tangent to all three spheres; the second plane can be written as the equation $ax + bx + cz = d$ for some real numbers $a,$ $b,$ $c,$ and $d.$  Find $\frac{c}{a}.$
The equation for the $xy$-plane is $z = 0.$

[asy]
import three;
import solids;

size(300);
currentprojection = perspective(-2,-2,3);

draw((2,17,0)--(17,2,0)--(-8,-29,0)--(-29,-8,0)--cycle);
draw(shift((4,7,5))*surface(sphere(5)),gray(0.8));
draw(shift((-2,5,4))*surface(sphere(4)),gray(0.8));
draw(shift((2,1,4))*surface(sphere(4)),gray(0.8));
draw((2,17,6)--(17,2,6)--(-8,-29,-1)--(-29,-8,-1)--cycle);
draw((0,-29,0)--(-29,0,0));

label("$x + y = -29$", (0,-29,0), E);
[/asy]

Let $\mathbf{a} = \begin{pmatrix} -2 \\ 5 \\ 4 \end{pmatrix},$ $\mathbf{b} = \begin{pmatrix} 2 \\ 1 \\ 4 \end{pmatrix},$ and $\mathbf{c} = \begin{pmatrix} 4 \\ 7 \\ 5 \end{pmatrix}.$  Then the normal vector to the plane passing through the centers of the spheres is
\[(\mathbf{b} - \mathbf{a}) \times (\mathbf{c} - \mathbf{a}) = \begin{pmatrix} 4 \\ -4 \\ 0 \end{pmatrix} \times \begin{pmatrix} 6 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} -4 \\ -4 \\ 32 \end{pmatrix}.\]We can scale this vector, and take $\begin{pmatrix} 1 \\ 1 \\ -8 \end{pmatrix}$ as the normal vector.  Thus, the equation of the plane is of the form $x + y - 8z = d.$  Substituting any of the centers, we find the equation of this plane is
\[x + y - 8z = -29.\]The intersection of this plane with the plane $z = 0$ is the line defined by
\[x + y = -29.\]The equation of any plane containing this line is then of the form
\[kx + ky + z = -29k.\]We want all three spheres to be tangent to this plane.  Then the distance between this plane and the center $(-2,5,4)$ should be 4.  From the formula for the distance between a point and a plane,
\[\frac{|-2k + 5k + 4 + 29k|}{\sqrt{k^2 + k^2 + 1}} = 4.\]Then $|32k + 4| = 4 \sqrt{2k^2 + 1},$ so $|8k + 1| = \sqrt{2k^2 + 1}.$  Squaring, we get $64k^2 + 16k + 1 = 2k^2 + 1,$ which simplifies to
\[62k^2 + 16k = 2k(31k + 8) = 0.\]The solutions are $k = 0$ and $k = -\frac{8}{31}.$  The solution $k = 0$ corresponds to the plane $z = 0,$ so the other plane corresponds to $k = -\frac{8}{31},$ which gives us the equation
\[-\frac{8}{31} x - \frac{8}{31} y + z = 29 \cdot \frac{8}{31}.\]Thus, $\frac{c}{a} = \boxed{-\frac{31}{8}}.$