One sphere is centered at $(3,-5,7)$ with radius $5 \sqrt{5}.$  A second sphere is centered at $(0,1,1)$ with radius $2 \sqrt{17}.$  The two spheres intersect in a circle.  Find the radius of this circle.
Let $A = (3,-5,7),$ the center of the first sphere, and let $B = (0,1,1),$ the center of the second sphere.  We can compute that $AB = 9.$

Let $C$ be a point on the intersection of both spheres, so $AC = 5 \sqrt{5}$ and $BC = 2 \sqrt{17}.$

[asy]
unitsize(0.3 cm);

pair A, B, C;

A = (0,0);
B = (9,0);
C = intersectionpoint(arc(A,5*sqrt(5),0,180),arc(B,2*sqrt(17),0,180));

draw(A--B--C--cycle);
draw(Circle(A,5*sqrt(5)));
draw(Circle(B,2*sqrt(17)));

label("$A$", A, W);
label("$B$", B, S);
label("$C$", C, N);
label("$9$", (A + B)/2, S, red);
label("$5 \sqrt{5}$", (A + C)/2, NW, red, UnFill);
label("$2 \sqrt{17}$", (B + C)/2, E, red, UnFill);
[/asy]

By Heron's formula, we can compute that $[ABC] = 3 \sqrt{149}.$

Let $D$ be the foot of the perpendicular from $C$ to $\overline{AB}.$

[asy]
unitsize(0.3 cm);

pair A, B, C, D;

A = (0,0);
B = (9,0);
C = intersectionpoint(arc(A,5*sqrt(5),0,180),arc(B,2*sqrt(17),0,180));
D = (C.x,0);

draw(A--B--C--cycle);
draw(C--D);

label("$A$", A, W);
label("$B$", B, S);
label("$C$", C, N);
label("$D$", D, S);
[/asy]

Then the intersection of both spheres is the circle centered at $D$ with radius $CD.$  Thus,
\[CD = \frac{2 [ABC]}{AB} = \frac{6 \sqrt{149}}{9} = \boxed{\frac{2 \sqrt{149}}{3}}.\]