Let position vectors of point A,B and C of triangle ABC represents be `hati+hatj+2hatk, hati+2hatj+hatk` and `2hati+hatj+hatk`. Let `l_(1)+l_(2)` and `l_(3)` be the length of perpendicular drawn from the orthocenter 'O' on the sides AB, BC and CA, then `(l_(1)+l_(2)+l_(3))` equals

Options

A. `2/sqrt(6)`
B. `3/sqrt(6)`
C. `sqrt(6)/2`
D. `sqrt(6)/3`

Explanation

To solve the problem, we need to find the lengths of the perpendiculars from the orthocenter \( O \) of triangle \( ABC \) to the sides \( AB \), \( BC \), and \( CA \). We will denote these lengths as \( l_1 \), \( l_2 \), and \( l_3 \) respectively, and we need 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 extract the coordinates:
- \( A(1, 1, 2) \)
- \( B(1, 2, 1) \)
- \( C(2, 1, 1) \)

### Step 3: Calculate the lengths of the sides of triangle ABC
Using the distance formula, we calculate the lengths of the sides:
1. **Length \( AB \)**:
\[ AB = \sqrt{(1 - 1)^2 + (2 - 1)^2 + (1 - 2)^2} = \sqrt{0 + 1 + 1} = \sqrt{2} \]

2. **Length \( BC \)**:
\[ BC = \sqrt{(2 - 1)^2 + (1 - 2)^2 + (1 - 1)^2} = \sqrt{1 + 1 + 0} = \sqrt{2} \]

3. **Length \( CA \)**:
\[ CA = \sqrt{(2 - 1)^2 + (1 - 1)^2 + (1 - 2)^2} = \sqrt{1 + 0 + 1} = \sqrt{2} \]

### Step 4: Determine the type of triangle
Since \( AB = BC = CA = \sqrt{2} \), triangle \( ABC \) is an equilateral triangle.

### Step 5: Calculate the area \( \Delta \) of triangle ABC
For an equilateral triangle, the area can be calculated using the formula:
\[ \Delta = \frac{\sqrt{3}}{4} s^2 \]
where \( s \) is the length of a side. Here, \( s = \sqrt{2} \):
\[ \Delta = \frac{\sqrt{3}}{4} (\sqrt{2})^2 = \frac{\sqrt{3}}{4} \cdot 2 = \frac{\sqrt{3}}{2} \]

### Step 6: Calculate the semi-perimeter \( S \)
The semi-perimeter \( S \) is given by:
\[ S = \frac{AB + BC + CA}{2} = \frac{\sqrt{2} + \sqrt{2} + \sqrt{2}}{2} = \frac{3\sqrt{2}}{2} \]

### Step 7: Calculate the inradius \( r \)
The inradius \( r \) is given by:
\[ r = \frac{\Delta}{S} = \frac{\frac{\sqrt{3}}{2}}{\frac{3\sqrt{2}}{2}} = \frac{\sqrt{3}}{3\sqrt{2}} = \frac{\sqrt{3}}{3\sqrt{2}} = \frac{\sqrt{3}}{3\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6}}{6} \]

### Step 8: Calculate \( l_1 + l_2 + l_3 \)
Since \( l_1 = l_2 = l_3 = r \):
\[ l_1 + l_2 + l_3 = 3r = 3 \cdot \frac{\sqrt{6}}{6} = \frac{3\sqrt{6}}{6} = \frac{\sqrt{6}}{2} \]

### Final Answer
Thus, the value of \( l_1 + l_2 + l_3 \) is \( \frac{\sqrt{6}}{2} \).

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