Is it long?

A solution using similarity and Pythagorean Theorem:

Let the midpoint of BC be M. Let the two other points of tangency of the incircle be J and L. Let K be the midpoint of the segment JL.

Note that G,I and M all lie on the perpendicular bisector of $\triangle ABC$ , so they are collinear.

Let the inradius be r. $\frac {1}{2}r(14+14+20)=Area=\sqrt{24*10*10*4}$ (using Heron's formula)

Hence $r=\frac {5}{3}\sqrt{6}$ .

It is known that the centroid divides a median into 2 segments of ratio $1:2$ .

$AM=\sqrt{14^2-10^2}=4\sqrt{6}$

Also, $\triangle ABC \sim \triangle AJL$ and $JB=BM=10$ , so $JK= \frac {4}{14}\sqrt{14^4-10^2}=\frac {8}{7}\sqrt{6}$

Thus $GM=\frac {2}{3}*(4\sqrt{6}-\frac {8}{7}\sqrt{6})=\frac {40}{21}\sqrt{6}$

$\Rightarrow IG=GM-IM=\frac {40}{21}\sqrt{6}-\frac {5}{3}\sqrt{6}=\frac {5\sqrt{6}}{21}$ $\Rightarrow a+b+c=5+6+21=\boxed{32}$

$\text{ Let M be the midpoint of BC.} \\ \therefore ~ AM ~ by ~ Pythagoras~ is~ 4\sqrt6. ~~~ M~ is ~0ne~ vertex ~ and ~ let ~ T_1, ~ T_2\\ be~other ~ two~ vertices ~ of~ the~ \Delta ~formed ~ by ~tangency~ points.\\ Inradius,~ r=\dfrac{\frac 1 2 *20*4\sqrt6}{\frac 1 2 *(14+14+20)}=\dfrac 5 3 \sqrt6.\\ Sin\dfrac A 2=\dfrac 5 7 . ~~~~~~~\color{#3D99F6}{-T_1G=+T_2G= r*Sin\dfrac A 2=\dfrac 5 7 *r}.\\ \text{Let I be the origin of the rectangular coordinate system with AM along X=0.}\\ \therefore I(0,0) , ~~M(0,-r), ~~ T_1(-p,\dfrac 5 7 *r), ~T_2(+p,\dfrac 5 7 *r), ~\\ \therefore ~G \left( \dfrac 0 3,\dfrac{r(-1+\dfrac 5 7+\dfrac 5 7 )} 3 \right )=\left( 0 ,\dfrac {5\sqrt6}{21} \right) .\\ \therefore ~IG=\dfrac {5\sqrt6}{21}=\dfrac {a\sqrt b}c.\\ \therefore ~~a+b+c=\Large ~~~~~\color{#D61F06}{32}$