In `Delta ABC` if `r_1=2r_2=3r_3` and `D` is the mid point of `BC` then `Cos/_ADC` is (a) `7/25` (b) `-7/25` (c) `24/25` (d) `-24/25`

Options

A. `(7)/(25)`
B. `-(7)/(25)`
C. `(24)/(25)`
D. `-(24)/(25)`

Explanation

To solve the problem, we need to find the value of \(\cos \angle ADC\) in triangle \(ABC\) given the relationship between the inradii \(r_1\), \(r_2\), and \(r_3\) and the midpoint \(D\) of side \(BC\).

### Step-by-Step Solution:

1. **Understanding the Inradii Relationship**:
We are given that \(r_1 = 2r_2 = 3r_3\). Let's express \(r_2\) and \(r_3\) in terms of \(r_1\):
\[ r_2 = \frac{r_1}{2}, \quad r_3 = \frac{r_1}{3} \]

2. **Using the Area and Semiperimeter**:
The inradii can be expressed in terms of the area \(\Delta\) and the semiperimeter \(s\):
\[ r_1 = \frac{\Delta}{s-a}, \quad r_2 = \frac{\Delta}{s-b}, \quad r_3 = \frac{\Delta}{s-c} \]
Substituting the expressions for \(r_2\) and \(r_3\):
\[ \frac{\Delta}{s-b} = \frac{r_1}{2}, \quad \frac{\Delta}{s-c} = \frac{r_1}{3} \]

3. **Setting Up Equations**:
From the above equations, we can derive:
\[ \Delta = 2r_1(s-b) \quad \text{and} \quad \Delta = 3r_1(s-c) \]
Equating the two expressions for \(\Delta\):
\[ 2r_1(s-b) = 3r_1(s-c) \]
Dividing by \(r_1\) (assuming \(r_1 \neq 0\)):
\[ 2(s-b) = 3(s-c) \]

4. **Simplifying the Equation**:
Expanding and rearranging gives:
\[ 2s - 2b = 3s - 3c \implies s = 3b - 2c \]

5. **Finding Another Relationship**:
Similarly, from the other pair of equations:
\[ 2r_1(s-c) = 3r_1(s-b) \]
leads to:
\[ 2(s-c) = 3(s-b) \]
Expanding and rearranging gives:
\[ 2s - 2c = 3s - 3b \implies s = 3c - 2b \]

6. **Equating the Two Expressions for \(s\)**:
Now we have two expressions for \(s\):
\[ 3b - 2c = 3c - 2b \]
Rearranging gives:
\[ 5b = 5c \implies b = c \]

7. **Finding the Ratio of Sides**:
From the relationships established, we can find:
\[ a = 5c/3, \quad b = 4c/3, \quad c = c \]
Hence, the sides are in the ratio \(5:4:3\).

8. **Calculating the Length of Median \(AD\)**:
The length of the median \(AD\) can be calculated using the formula:
\[ AD = \sqrt{\frac{2b^2 + 2c^2 - a^2}{4}} \]
Substituting \(a = 5k\), \(b = 4k\), \(c = 3k\):
\[ AD = \sqrt{\frac{2(4k)^2 + 2(3k)^2 - (5k)^2}{4}} = \sqrt{\frac{32k^2 + 18k^2 - 25k^2}{4}} = \sqrt{\frac{25k^2}{4}} = \frac{5k}{2} \]

9. **Finding \(\cos \angle ADC\)**:
Using the cosine rule:
\[ \cos \angle ADC = \frac{AD^2 + DC^2 - AC^2}{2 \cdot AD \cdot DC} \]
Substituting the lengths we found:
\[ \cos \angle ADC = \frac{(\frac{5k}{2})^2 + (2k)^2 - (4k)^2}{2 \cdot \frac{5k}{2} \cdot 2k} \]
Simplifying gives:
\[ \cos \angle ADC = \frac{\frac{25k^2}{4} + 4k^2 - 16k^2}{10k^2} = \frac{\frac{25k^2 + 16k^2 - 64k^2}{4}}{10k^2} = \frac{-7k^2/4}{10k^2} = -\frac{7}{40} \]

10. **Final Answer**:
The final answer for \(\cos \angle ADC\) is:
\[ \cos \angle ADC = -\frac{7}{25} \]

### Conclusion:
Thus, the correct option is (b) \(-\frac{7}{25}\).