Key Concepts and Formulas

1. Vector Magnitude: The magnitude of a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ is $||\vec{v}|| = \sqrt{x^2 + y^2 + z^2}$. The distance between two points with position vectors $\vec{a}$ and $\vec{b}$ is $||\vec{b} - \vec{a}||$.
2. Centroid of a Triangle: The position vector of the centroid $\vec{G}$ of a triangle with vertices at $\vec{a}, \vec{b}, \vec{c}$ is $\vec{G} = \frac{\vec{a}+\vec{b}+\vec{c}}{3}$.
3. Properties of an Equilateral Triangle: In an equilateral triangle, the orthocenter, centroid, incenter, and circumcenter coincide. The perpendicular distance from the orthocenter to each side is equal to the inradius ($r$). The inradius of an equilateral triangle with side length $a$ is $r = \frac{a}{2\sqrt{3}}$.

Step-by-Step Solution

Step 1: Calculate the lengths of the sides of the triangle. To determine the type of triangle, we first compute the lengths of its sides using the given position vectors: $\vec{a} = 2 \hat{i}+2 \hat{j}+\hat{k}$ $\vec{b} = \hat{i}+2 \hat{j}+2 \hat{k}$ $\vec{c} = 2 \hat{i}+\hat{j}+2 \hat{k}$

Length of AB: $||\vec{AB}|| = ||\vec{b} - \vec{a}|| = ||(\hat{i}+2\hat{j}+2\hat{k}) - (2\hat{i}+2\hat{j}+\hat{k})|| = ||-\hat{i} + 0\hat{j} + \hat{k}|| = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{2}$

Length of BC: $||\vec{BC}|| = ||\vec{c} - \vec{b}|| = ||(2\hat{i}+\hat{j}+2\hat{k}) - (\hat{i}+2\hat{j}+2\hat{k})|| = ||\hat{i} - \hat{j} + 0\hat{k}|| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2}$

Length of CA: $||\vec{CA}|| = ||\vec{a} - \vec{c}|| = ||(2\hat{i}+2\hat{j}+\hat{k}) - (2\hat{i}+\hat{j}+2\hat{k})|| = ||0\hat{i} + \hat{j} - \hat{k}|| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}$ Since all sides have equal length ($\sqrt{2}$), triangle ABC is an equilateral triangle.

Step 2: Determine the position vector of the orthocenter. For an equilateral triangle, the orthocenter coincides with the centroid. We find the position vector of the centroid using the formula: $\vec{h} = \frac{\vec{a}+\vec{b}+\vec{c}}{3} = \frac{(2\hat{i}+2\hat{j}+\hat{k}) + (\hat{i}+2\hat{j}+2\hat{k}) + (2\hat{i}+\hat{j}+2\hat{k})}{3}$ $\vec{h} = \frac{(2+1+2)\hat{i} + (2+2+1)\hat{j} + (1+2+2)\hat{k}}{3} = \frac{5\hat{i}+5\hat{j}+5\hat{k}}{3}$

Step 3: Calculate the lengths of the perpendiculars from the orthocenter to the sides. Let $l_1, l_2, l_3$ be the lengths of the perpendiculars from the orthocenter to sides AB, BC, and CA, respectively. Since the triangle is equilateral and the orthocenter is also the incenter, these lengths are all equal to the inradius ($r$) of the triangle. The side length $a = \sqrt{2}$. The inradius $r$ of an equilateral triangle is given by $r = \frac{a}{2\sqrt{3}}$. $r = \frac{\sqrt{2}}{2\sqrt{3}} = \frac{1}{\sqrt{2}\sqrt{3}} = \frac{1}{\sqrt{6}}$ Therefore, $l_1 = l_2 = l_3 = \frac{1}{\sqrt{6}}$.

Step 4: Calculate $l_1^2+l_2^2+l_3^2$. Now, we sum the squares of these lengths: $l_1^2+l_2^2+l_3^2 = \left(\frac{1}{\sqrt{6}}\right)^2 + \left(\frac{1}{\sqrt{6}}\right)^2 + \left(\frac{1}{\sqrt{6}}\right)^2$ $l_1^2+l_2^2+l_3^2 = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$

Common Mistakes & Tips

• Identify Triangle Type Early: Always calculate side lengths first. Recognizing the triangle as equilateral is key to simplifying the orthocenter and perpendicular distance calculations.
• Orthocenter Properties: For an equilateral triangle, the orthocenter, centroid, and incenter are the same point. This simplifies finding the orthocenter's location and the perpendicular distances.
• Inradius Formula: Recall the formula for the inradius of an equilateral triangle ($r = a/(2\sqrt{3})$) to quickly find the lengths of the perpendiculars.

Summary The problem involves calculating the lengths of the sides of the triangle formed by the given position vectors. Upon finding that the triangle is equilateral, we leverage the property that its orthocenter coincides with its centroid and incenter. The lengths of the perpendiculars from the orthocenter to the sides are equal to the inradius. By calculating the side length and then the inradius, we find these lengths. Finally, summing the squares of these lengths yields the answer.

The final answer is $\boxed{\frac{1}{2}}$, which corresponds to option (D).