$PB = QB = radius = r$ (Let us say)

$AP = \frac{1}{2}PB$
$AP = \frac{r}{2}$

$AB = AP + PB = \frac{r}{2} + r = \frac{3r}{2}$

$CQ = \frac{2}{7}QB$
$CQ = \frac{2r}{7}$

$BC = CQ + QB = \frac{2r}{7} + r = \frac{9r}{7}$

Since, ABC is a Right angled triangle, we can apply Pythagoras theorem;

$AC^2 = AB^2 + BC^2$
$100^2 = (\frac{3r}{2})^2 + (\frac{9r}{7})^2$

$100^2 = \frac{9r^2}{4} + \frac{81r^2}{49}$

$(4)(49)100^2 = (9r^2)(49) + (81r^2)(4)$

$(4)(49)100^2 = 765r^2$

$r^2 = \frac{(4)(49)100^2}{765}$

$r = \sqrt{\frac{(4)(49)100^2}{765}} = \frac{1400}{\sqrt{765}}$

$r = \frac{1400}{3\sqrt{85}}$

Since, we have $\sqrt{85}$ in option A - it must be the our Answer!