If $A = (3,4)$ and $C = (7,10)$ are opposite vertices of a rectangle $ABCD,$ then vertices $A, B, C,$ and $D$ must lie on the circle
\[x^2 + y^2 - px - qy + s = 0.\]Compute the ordered triple of real numbers $(p,q,s).$
Since $\angle ABC = \angle ADC = 90^\circ,$ $\overline{AC}$ must be a diameter of this circle.

[asy]
unitsize (0.8 cm);

pair A, B, C, D, O;

A = (3,4);
C = (7,10);
O = (A + C)/2;
B = O + abs(O - A)*dir(170);
D = O + abs(O - A)*dir(350);

draw(Circle(O,abs(O - A)));
draw(A--C);
draw(B--D);
draw(A--B--C--D--cycle);

label("$A$", A, SW);
label("$B$", B, W);
label("$C$", C, NE);
label("$D$", D, E);

dot(O);
[/asy]

The center of the circle is then $(5,7)$ (the mid-point of $A$ and $C$), and its radius is $\sqrt{(5 - 3)^2 + (7 - 4)^2} = \sqrt{13},$ so its equation is
\[(x - 5)^2 + (y - 7)^2 = 13.\]This simplifies to $x^2 + y^2 - 10x - 14y + 61 = 0,$ so $(p,q,s) = \boxed{(10,14,61)}.$