1. Write the vertices as points

Let $A=(2,2,1),\quad B=(1,2,2),\quad C=(2,1,2).$

We need the distances from the orthocenter $H$ of triangle $ABC$ to the sides $AB,BC,CA$.

2. Find the side lengths

Using distance formula:

$AB^2=(2-1)^2+(2-2)^2+(1-2)^2=1+0+1=2$ $BC^2=(1-2)^2+(2-1)^2+(2-2)^2=1+1+0=2$ $CA^2=(2-2)^2+(1-2)^2+(2-1)^2=0+1+1=2$

Hence $AB=BC=CA=\sqrt{2}.$

So the triangle is equilateral.

3. Use the orthocenter property in an equilateral triangle

In an equilateral triangle, the orthocenter, centroid, circumcenter and incenter all coincide. Therefore the perpendicular distances from the orthocenter to all three sides are equal to the inradius $r$.

So $l_1=l_2=l_3=r.$ Hence $l_1^2+l_2^2+l_3^2=3r^2.$

4. Find the inradius of the equilateral triangle

For an equilateral triangle of side $a$, $r=\frac{a\sqrt{3}}{6}.$

Here $a=\sqrt{2}$, so $r=\frac{\sqrt{2}\sqrt{3}}{6}=\frac{\sqrt{6}}{6}.$

Thus $r^2=\frac{6}{36}=\frac{1}{6}.$

Therefore, $l_1^2+l_2^2+l_3^2=3\cdot \frac{1}{6}=\frac{1}{2}.$

5. Check options

$\frac{1}{2}$ matches Option D.

Final Answer: $\boxed{\frac{1}{2}}$