Answer is: 164
By given data
\(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{CB}}\)
Let pv of \(\overrightarrow{\mathrm{A}}\) are \(\overrightarrow{\mathrm{O}}\) then
\(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}\)
i.e. pv of \(\overrightarrow{\mathrm{B}}-=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{C}}\)
i.e. pv of \(\vec{C}=-(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})\)
Now pv of centroid
\((\overrightarrow{\mathrm{G}})-=\frac{\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}}{3}=\frac{\overrightarrow{0}+(2,-1,1)+(-1,3,5)}{3}\)
\(\overrightarrow{\mathrm{G}}=\frac{1}{3}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})\)
Now \(\overrightarrow{\mathrm{AG}}=\frac{1}{3}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})\)
\(\Rightarrow|\overrightarrow{\mathrm{AG}}|^{2}=\frac{1}{9} \times 41\)
\(\overrightarrow{\mathrm{BG}}=\left(\frac{1}{3}-2\right) \hat{\mathrm{i}}+\left(\frac{2}{3}+1\right) \hat{\mathrm{j}}+(2-1) \hat{\mathrm{k}}\)
\(\Rightarrow|\overrightarrow{\mathrm{BG}}|^{2}=\frac{59}{9}\)
\(\overrightarrow{\mathrm{CG}}=\left(\frac{1}{3}+1\right) \hat{\mathrm{i}}+\left(\frac{2}{3}-3\right) \hat{\mathrm{j}}+(2-5) \hat{\mathrm{k}}\)
\(\Rightarrow|\overrightarrow{\mathrm{CG}}|^{2}=\frac{146}{9}\)
Now
\(6\left[|\overrightarrow{\mathrm{AG}}|^{2}+|\overrightarrow{\mathrm{BG}}|^{2}+|\overrightarrow{\mathrm{CG}}|^{2}\right]=6 \times\left[\frac{41}{9}+\frac{59}{9}+\frac{146}{9}\right]\)
\(=6 \times \frac{246}{9}=164\)
