I rocks everywhere

Here is a solution without the use of Trignometry. It is well known that $I$ is the orthocenter of $I_1I_2I_3$ , and $\odot ABC$ is essentially the nice point circle of $I_1I_2I_3$ . Therefore we can reword the problem the following way: Given a triangle $ABC$ with orthocenter $H$ and circumradius $2$ , find $\sum_{cyc} AH^2+BC^2$ . (Note that half of the circumradius is the radius of the nine point circle.

We will show that $AH^2+BC^2$ is actually constant, given the circumradius is fixed. To see this, let $O$ denote the circumcenter of $ABC$ , $M$ is the midpoint of $BC$ . It should be well known that $2OM=AH$ , therefore $AH^2+BC^2=4OM^2+4MC^2=4OC^2=4(2)^2=16$ . Hence our answer is $16*3=48$ .

As per standard trigonometric results we know the following things:

$II_1=4R\sin\dfrac{A}{2} \\ II_2=4R\sin\dfrac{B}{2} \\ II_1=4R\sin\dfrac{C}{2} \\ I_1I_2=4R\cos\dfrac{C}{2} \\ I_2I_3=4R\cos\dfrac{A}{2} \\I_1I_3=4R\cos\dfrac{B}{2} \\ II_1^2+I_2I_3^2= 16R^2\sin^2\dfrac{A}{2} + 16R^2\cos^2\dfrac{A}{2}= 16R^2\left(\sin^2\dfrac{A}{2} + \cos^2\dfrac{A}{2}\right) \\ = 16 \\ II_2^2+I_1I_3^2= 16R^2\sin^2\dfrac{B}{2} + 16R^2\cos^2\dfrac{B}{2}= 16R^2\left(\sin^2\dfrac{B}{2} + \cos^2\dfrac{B}{2}\right) \\ = 16 \\ II_3^2+I_1I_2^2= 16R^2\sin^2\dfrac{C}{2} + 16R^2\cos^2\dfrac{C}{2}= 16R^2\left(\sin^2\dfrac{C}{2} + \cos^2\dfrac{C}{2}\right) \\ = 16$

So the required sum $\large =16+16+16=\boxed{48}$