Let the position vectors of the vertices A, B and C of a triangle...

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

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

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

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

So, $∆ ABC$ is equilateral.

Therefore, orthocentre and centroid will be same.

$⇒ G ≡ \left(\dfrac{2 + 1 + 2}{3}, \dfrac{2 + 2 + 1}{3}, \dfrac{1 + 2 + 2}{3}\right)$

$⇒ G ≡ \left(\dfrac{5}{3}, \dfrac{5}{3}, \dfrac{5}{3}\right)$

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

$⇒ {l}_{1} = \sqrt{{\left(\dfrac{3}{2} - \dfrac{5}{3}\right)}^{2} + {\left(2 - \dfrac{5}{3}\right)}^{2} + {\left(\dfrac{3}{2} - \dfrac{5}{3}\right)}^{2}}$

$⇒ {l}_{1} = \sqrt{\dfrac{1}{36} + \dfrac{1}{9} + \dfrac{1}{36}}$

$⇒ {l}_{1} = \sqrt{\dfrac{1}{6}}$

Similarly, ${l}_{2} = {l}_{3} = \sqrt{\dfrac{1}{6}}$

$⇒ {\left({l}_{1}\right)}^{2} + {\left({l}_{2}\right)}^{2} + {\left({l}_{3}\right)}^{2} = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6}$

$⇒ {\left({l}_{1}\right)}^{2} + {\left({l}_{2}\right)}^{2} + {\left({l}_{3}\right)}^{2} = \dfrac{1}{2}$