1. Let the three side vectors of triangle $ABC$ be $\vec{AB},\ \vec{BC},\ \vec{CA}$ with magnitudes corresponding to the given vectors: $\vec{u}=2\hat i-\hat j+\hat k,$ $\vec{v}=\hat i-3\hat j-5\hat k,$ $\vec{w}=3\hat i-4\hat j-4\hat k.$

   First, check consistency: $\vec{u}+\vec{v}=(2+1)\hat i+(-1-3)\hat j+(1-5)\hat k=3\hat i-4\hat j-4\hat k=\vec{w}.$ So these can indeed be the three sides of a triangle, with $\vec{AB}=\vec{u},\quad \vec{BC}=\vec{v},\quad \vec{AC}=\vec{w}=\vec{u}+\vec{v}.$

2. Use the centroid identity: For a triangle with centroid $G$, $AG^2+BG^2+CG^2=\frac{1}{3}(AB^2+BC^2+CA^2).$ Therefore, $6(AG^2+BG^2+CG^2)=6\cdot \frac{1}{3}(AB^2+BC^2+CA^2)=2(AB^2+BC^2+CA^2).$

3. Compute the squared lengths of the three sides.

   For $\vec{u}=2\hat i-\hat j+\hat k$: $|\vec{u}|^2=2^2+(-1)^2+1^2=4+1+1=6.$

   For $\vec{v}=\hat i-3\hat j-5\hat k$: $|\vec{v}|^2=1^2+(-3)^2+(-5)^2=1+9+25=35.$

   For $\vec{w}=3\hat i-4\hat j-4\hat k$: $|\vec{w}|^2=3^2+(-4)^2+(-4)^2=9+16+16=41.$

4. Add them: $AB^2+BC^2+CA^2=6+35+41=82.$

5. Hence, $6(AG^2+BG^2+CG^2)=2\times 82=164.$

So the required integer is $\boxed{164}.$