To solve the problem, we need to find the lengths of the perpendiculars from the orthocenter \( O \) of triangle \( ABC \) to its sides \( AB \), \( BC \), and \( CA \). We will denote these lengths as \( l_1 \), \( l_2 \), and \( l_3 \) respectively. The goal is to compute \( l_1 + l_2 + l_3 \).
### Step 1: Identify the position vectors of points A, B, and C
The position vectors are given as:
- \( \vec{A} = \hat{i} + \hat{j} + 2\hat{k} \)
- \( \vec{B} = \hat{i} + 2\hat{j} + \hat{k} \)
- \( \vec{C} = 2\hat{i} + \hat{j} + \hat{k} \)
### Step 2: Convert position vectors to coordinates
From the position vectors, we can express the coordinates of points A, B, and C:
- \( A(1, 1, 2) \)
- \( B(1, 2, 1) \)
- \( C(2, 1, 1) \)
### Step 3: Calculate the lengths of the sides of triangle ABC
Using the distance formula, we calculate the lengths of sides \( AB \), \( BC \), and \( CA \).
1. **Length of side \( AB \)**:
\[
AB = \sqrt{(1-1)^2 + (2-1)^2 + (1-2)^2} = \sqrt{0 + 1 + 1} = \sqrt{2}
\]
2. **Length of side \( BC \)**:
\[
BC = \sqrt{(2-1)^2 + (1-2)^2 + (1-1)^2} = \sqrt{1 + 1 + 0} = \sqrt{2}
\]
3. **Length of side \( CA \)**:
\[
CA = \sqrt{(2-1)^2 + (1-1)^2 + (1-2)^2} = \sqrt{1 + 0 + 1} = \sqrt{2}
\]
### Step 4: Determine the area of triangle ABC
Since \( AB = BC = CA = \sqrt{2} \), triangle \( ABC \) is an equilateral triangle. The area \( \Delta \) of an equilateral triangle with side length \( a \) is given by:
\[
\Delta = \frac{\sqrt{3}}{4} a^2
\]
Substituting \( a = \sqrt{2} \):
\[
\Delta = \frac{\sqrt{3}}{4} (\sqrt{2})^2 = \frac{\sqrt{3}}{4} \cdot 2 = \frac{\sqrt{3}}{2}
\]
### Step 5: Calculate the semi-perimeter \( s \)
The semi-perimeter \( s \) of triangle \( ABC \) is:
\[
s = \frac{AB + BC + CA}{2} = \frac{\sqrt{2} + \sqrt{2} + \sqrt{2}}{2} = \frac{3\sqrt{2}}{2}
\]
### Step 6: Calculate the circumradius \( R \)
For an equilateral triangle, the circumradius \( R \) is given by:
\[
R = \frac{a}{\sqrt{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3}
\]
### Step 7: Calculate the lengths of the perpendiculars \( l_1, l_2, l_3 \)
For an equilateral triangle, the lengths of the perpendiculars from the orthocenter to each side are equal and can be calculated as:
\[
l = \frac{2\Delta}{a}
\]
Substituting \( \Delta = \frac{\sqrt{3}}{2} \) and \( a = \sqrt{2} \):
\[
l = \frac{2 \cdot \frac{\sqrt{3}}{2}}{\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2}
\]
### Step 8: Calculate \( l_1 + l_2 + l_3 \)
Since \( l_1 = l_2 = l_3 = \frac{\sqrt{6}}{2} \):
\[
l_1 + l_2 + l_3 = 3l = 3 \cdot \frac{\sqrt{6}}{2} = \frac{3\sqrt{6}}{2}
\]
### Final Answer
Thus, the value of \( l_1 + l_2 + l_3 \) is:
\[
\frac{3\sqrt{6}}{2}
\]
