Problem 99

Let the center of circle C be $(a, b)$, and its radius be $r$.
Since the center of circle C is on the line $l_1: x-y-1=0$,
we have $a-b-1=0$,
Since circle C is tangent to the line $l_2: 4x+3y+14=0$,
we have $r= \frac {|4a+3b+14|}{5}$,
Since the chord intercepted by the line $l_3: 3x+4y+10=0$ on circle C has a length of 6,
we have $\frac {|3a+4b+10|}{5}= \sqrt {r^{2}-9}$.
Therefore, $\frac {(4a+3b+14)^{2}}{25} - \frac {(3a+4b+10)^{2}}{25} = 9$.
That is, $\frac {(a-b+4)(7a+7b+24)}{25} = 9$.
Since $a-b=1$,
we have $\frac {5(7a+7b+24)}{25} = 9$,
thus $a+b=3$.
Solving $\begin{cases} a-b=1 \\ a+b=3 \end{cases}$ yields $\begin{cases} a=2 \\ b=1 \end{cases}$.
Therefore, the equation of circle C is $\boxed{(x-2)^2+(y-1)^2=25}$.