If $\left(0,\frac{3}{4}\right)$ is the radical centre of the circles $S\equiv x^2+y^2+\alpha x+6y=0,S^{\text{'}}\equiv x^2+y^2+2\alpha x+\alpha y+6=0$ and $S^{"}\equiv x^2+y^2+6\alpha x-\alpha y+3=0$ then the distance between the radical centre and the centre of the circle $S^{\text{'}}=0$ is

Given: 

$S\equiv x^2+y^2+\alpha x+6y=0$

$S^{\text{'}}\equiv x^2+y^2+2\alpha x+\alpha y+6=0$

$S^{"}\equiv x^2+y^2+6\alpha x-\alpha y+3=0$

Radical axis of $S$ and $S^{\text{'}}$ is

$S-S^{\text{'}}=0$

$\Rightarrow \alpha x+\left(\alpha -6\right)y+6=0 .....\left(\mathrm{i}\right)$

Radical axis of $S^{\text{'}}$ and $S^{"}$ is

$S^{\text{'}}-S^{"}=0$

$\Rightarrow 4\alpha -2\alpha y-3=0 ....\left(\mathrm{ii}\right)$

Solving $\left(\mathrm{i}\right) \& \left(\mathrm{ii}\right)$, we get

$y=\frac{9}{8-2\alpha }$

Hence,

$\frac{9}{8-2\alpha }=\frac{3}{4}$

$\Rightarrow 36=24-6\alpha$

$\Rightarrow \alpha =-2$

So,

$S^{\text{'}}\equiv x^2+y^2-4x-2y+6=0$

Centre $C_1\equiv \left(2,1\right)$

Required distance

$=\sqrt{{\left(2-0\right)}^2+{\left(1-\frac{3}{4}\right)}^2}$

$=\sqrt{4+\frac{1}{16}}=\frac{\sqrt{65}}{4}$ units.