Let the three sides of a triangle ABC be given by the vectors $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $3 \hat{i}-4 \hat{j}-4 \hat{k}$. Let $G$ be the centroid of the triangle $A B C$. Then $6\left(|\overrightarrow{\mathrm{AG}}|^2+|\overrightarrow{\mathrm{BG}}|^2+|\overrightarrow{\mathrm{CG}}|^2\right)$ is equal to __________.

<p>$$\begin{aligned} &\text { Assuming Vertex } A \text { to be origin }\\ &\begin{aligned} & \vec{A}=\vec{a}_1=\overrightarrow{0} \\ & \vec{B}=\vec{a}_1+\vec{u}=\vec{u}=2 \hat{i}-\hat{j}+\hat{k} \\ & \vec{C}=\vec{a}_1+\vec{v}=\vec{v}=3 \hat{i}-4 \hat{j}-4 \hat{k} \end{aligned} \end{aligned}$$</p> <p>One solving</p> <p>$\vec{A}=\overrightarrow{0}, \vec{B}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{C}=3 \hat{i}-4 \hat{j}-4 \hat{k}$, are the position vector of vertices $A B$ and $C$ respectively.</p> <p>$$\begin{aligned} & \vec{G}=\frac{1}{3}(\vec{A}+\vec{B}+\vec{C})=\frac{1}{3}(\overrightarrow{0}+\vec{B}+\vec{C})=\frac{1}{3}(\vec{B}+\vec{C}) \\ & \Rightarrow \vec{G}=\frac{5}{3} \hat{i}-\frac{5}{3} \hat{j}-\hat{k} \\ & \overrightarrow{A G}=\vec{G}-\vec{A}=\vec{G} \\ & |\overrightarrow{A G}|^2=\left(\frac{5}{3}\right)^2+\left(\frac{5}{3}\right)^2+(1)^2=\frac{25}{9}+\frac{25}{9}+1=\frac{50}{9}+1=\frac{59}{9} \\ & \overrightarrow{B C}=\vec{G}-\vec{B} \\ & \vec{B}=2 \hat{i}-\hat{j}+\hat{k} \\ & |\overrightarrow{B G}|^2=\left(\frac{1}{3}\right)^3+\left(\frac{2}{3}\right)^2+4=\frac{1}{9}+\frac{4}{9}+4=\frac{5}{9}+4=\frac{41}{9} \\ & \overrightarrow{C G}=\vec{G}-\vec{C} \\ & \vec{C}=3 \hat{i}-4 \hat{j}-4 \hat{k} \\ & \left.\overrightarrow{C G}\right|^2=\left(\frac{4}{3}\right)^2+\left(\frac{7}{3}\right)^2+9=\frac{16}{9}+\frac{49}{9}+9=\frac{65}{9}+9=\frac{65}{9}+\frac{81}{9}=\frac{146}{9} \end{aligned}$$</p> <p>$6\left(|\overrightarrow{A G}|^2+|\overrightarrow{B G}|^2+|\overrightarrow{C G}|^2\right)=6 \cdot\left(\frac{59}{9}+\frac{41}{9}+\frac{146}{9}\right)=6 \cdot \frac{246}{9}=164$</p>