In `DeltaABC, R, r, r_(1), r_(2), r_(3)` denote the circumradius, inradius, the exradii opposite to the vertices A,B, C respectively. Given that `r_(1) :r_(2): r_(3) = 1: 2 : 3`
The greatest angle of the triangle is given by

Options

A. `cos^(-1) ((1)/(30))`
B. `cos^(-1)((1)/(3))`
C. `cos^(-1)((1)/(10))`
D. `cos^(-1) ((1)/(5))`

Explanation

To solve the problem step by step, we will use the given ratios of the exradii and the relationships between the sides and angles of triangle \( ABC \).

### Step-by-Step Solution

1. **Understanding the Given Ratios**:
We are given that the exradii \( r_1 : r_2 : r_3 = 1 : 2 : 3 \). This implies:
\[ r_1 = k, \quad r_2 = 2k, \quad r_3 = 3k \]
for some positive constant \( k \).

2. **Using the Relationship between Exradii and Semiperimeter**:
The exradii are related to the area \( \Delta \) and the semiperimeter \( S \) of the triangle as follows:
\[ r_1 = \frac{\Delta}{S - A}, \quad r_2 = \frac{\Delta}{S - B}, \quad r_3 = \frac{\Delta}{S - C} \]
Thus, we can write:
\[ \frac{\Delta}{S - A} : \frac{\Delta}{S - B} : \frac{\Delta}{S - C} = 1 : 2 : 3 \]
This simplifies to:
\[ \frac{1}{S - A} : \frac{1}{S - B} : \frac{1}{S - C} = 1 : 2 : 3 \]

3. **Setting Up the Equations**:
Let:
\[ S - A = 6k, \quad S - B = 3k, \quad S - C = 2k \]

4. **Finding the Semiperimeter**:
Adding these equations gives:
\[ (S - A) + (S - B) + (S - C) = 6k + 3k + 2k = 11k \]
This simplifies to:
\[ 3S - (A + B + C) = 11k \]
Since \( A + B + C = 2S \), we can substitute:
\[ 3S - 2S = 11k \implies S = 11k \]

5. **Finding the Angles**:
Now substituting \( S \) back into the equations for \( A, B, \) and \( C \):
\[ A = S - 6k = 11k - 6k = 5k \]
\[ B = S - 3k = 11k - 3k = 8k \]
\[ C = S - 2k = 11k - 2k = 9k \]

6. **Finding the Greatest Angle**:
The greatest angle in triangle \( ABC \) is \( C \). We can find \( \cos C \) using the cosine rule:
\[ \cos C = \frac{A^2 + B^2 - C^2}{2AB} \]
Substituting the values:
\[ \cos C = \frac{(5k)^2 + (8k)^2 - (9k)^2}{2 \cdot (5k)(8k)} \]
This simplifies to:
\[ \cos C = \frac{25k^2 + 64k^2 - 81k^2}{80k^2} = \frac{8k^2}{80k^2} = \frac{1}{10} \]

7. **Finding Angle \( C \)**:
Therefore, the angle \( C \) is:
\[ C = \cos^{-1}\left(\frac{1}{10}\right) \]

### Final Answer
The greatest angle of triangle \( ABC \) is \( C = \cos^{-1}\left(\frac{1}{10}\right) \).