1. We need to find $\left|\hat i \times (\vec a \times \hat i)\right|^2+\left|\hat j \times (\vec a \times \hat j)\right|^2+\left|\hat k \times (\vec a \times \hat k)\right|^2$ for $\vec a=2\hat i+\hat j+2\hat k.$

2. Use the vector triple product identity: $\vec p\times(\vec q\times \vec r)=\vec q(\vec p\cdot \vec r)-\vec r(\vec p\cdot \vec q).$

So, $\hat i\times(\vec a\times \hat i)=\vec a(\hat i\cdot \hat i)-\hat i(\hat i\cdot \vec a)=\vec a-a_x\hat i.$ Since $a_x=2$, $\hat i\times(\vec a\times \hat i)=(2\hat i+\hat j+2\hat k)-2\hat i=\hat j+2\hat k.$ Hence, $\left|\hat i\times(\vec a\times \hat i)\right|^2=1^2+2^2=5.$

3. Similarly, $\hat j\times(\vec a\times \hat j)=\vec a-a_y\hat j.$ Since $a_y=1$, $\hat j\times(\vec a\times \hat j)=(2\hat i+\hat j+2\hat k)-\hat j=2\hat i+2\hat k.$ Thus, $\left|\hat j\times(\vec a\times \hat j)\right|^2=2^2+2^2=8.$

4. Again, $\hat k\times(\vec a\times \hat k)=\vec a-a_z\hat k.$ Since $a_z=2$, $\hat k\times(\vec a\times \hat k)=(2\hat i+\hat j+2\hat k)-2\hat k=2\hat i+\hat j.$ Therefore, $\left|\hat k\times(\vec a\times \hat k)\right|^2=2^2+1^2=5.$

5. Add them: $5+8+5=18.$

Therefore, the required integer is $\boxed{18}.$