Given: $\mathrm{A}\left(2,2,1\right), \mathrm{B}\left(1,2,2\right) \text{and} \mathrm{C}\left(2,1,2\right)$ are vertices of $∆\mathrm{ABC}$.

$\Rightarrow \mathrm{AB}=\sqrt{{\left(2-1\right)}^2+{\left(2-2\right)}^2+{\left(1-2\right)}^2}=\sqrt{2}$

$\Rightarrow \mathrm{BC}=\sqrt{{\left(1-2\right)}^2+{\left(2-1\right)}^2+{\left(2-2\right)}^2}=\sqrt{2}$

$\Rightarrow \mathrm{CA}=\sqrt{{\left(2-2\right)}^2+{\left(2-1\right)}^2+{\left(1-2\right)}^2}=\sqrt{2}$

So, $∆\mathrm{ABC}$ is equilateral.

Therefore, orthocentre and centroid will be same.

$\Rightarrow \mathrm{G}\equiv \left(\frac{2+1+2}{3},\frac{2+2+1}{3},\frac{1+2+2}{3}\right)$

$\Rightarrow \mathrm{G}\equiv \left(\frac{5}{3},\frac{5}{3},\frac{5}{3}\right)$

Also, mid point of $\mathrm{AB}$ is $\mathrm{D}\left(\frac{3}{2},2,\frac{3}{2}\right)$

$\Rightarrow {\mathrm{l}}_1=\sqrt{{\left(\frac{3}{2}-\frac{5}{3}\right)}^2+{\left(2-\frac{5}{3}\right)}^2+{\left(\frac{3}{2}-\frac{5}{3}\right)}^2}$

$\Rightarrow {\mathrm{l}}_1=\sqrt{\frac{1}{36}+\frac{1}{9}+\frac{1}{36}}$

$\Rightarrow {\mathrm{l}}_1=\sqrt{\frac{1}{6}}$

Similarly, ${\mathrm{l}}_2={\mathrm{l}}_3=\sqrt{\frac{1}{6}}$

$\Rightarrow {\left({\mathrm{l}}_1\right)}^2+{\left({\mathrm{l}}_2\right)}^2+{\left({\mathrm{l}}_3\right)}^2=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$

$\Rightarrow {\left({\mathrm{l}}_1\right)}^2+{\left({\mathrm{l}}_2\right)}^2+{\left({\mathrm{l}}_3\right)}^2=\frac{1}{2}$