Let ABC be an acute angled triangle with orthocenter H.D, E, and F are the feet of perpendicular from A,B, and C, respectively, on opposite sides. Also, let R be the circumradius of `DeltaABC`. Given `AH.BH.CH = 3 and (AH)^(2) + (BH)^(2) + (CH)^(2) = 7`
Then answer the following
Value of `HD.HE.HF` is

Options

A. `(9)/(64R^(3))`
B. `(9)/(8R^(3))`
C. `(8)/(9R^(3))`
D. `(64)/(9R^(3))`

Explanation

To solve the problem, we need to find the value of \( HD \cdot HE \cdot HF \) given the conditions about the triangle \( ABC \) and its orthocenter \( H \).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
We know that:
- \( AH \cdot BH \cdot CH = 3 \)
- \( AH^2 + BH^2 + CH^2 = 7 \)

2. **Using the Relationships of the Orthocenter**:
For an acute triangle, the lengths from the vertices to the orthocenter can be expressed as:
\[ AH = 2R \cos A, \quad BH = 2R \cos B, \quad CH = 2R \cos C \]
where \( R \) is the circumradius of triangle \( ABC \).

3. **Substituting into the Given Conditions**:
Substitute \( AH, BH, CH \) into the first condition:
\[ (2R \cos A)(2R \cos B)(2R \cos C) = 3 \]
This simplifies to:
\[ 8R^3 \cos A \cos B \cos C = 3 \quad \text{(1)} \]

4. **Substituting into the Second Condition**:
Substitute into the second condition:
\[ (2R \cos A)^2 + (2R \cos B)^2 + (2R \cos C)^2 = 7 \]
This simplifies to:
\[ 4R^2 (\cos^2 A + \cos^2 B + \cos^2 C) = 7 \quad \text{(2)} \]

5. **Finding \( HD, HE, HF \)**:
The distances from the orthocenter to the feet of the altitudes are given by:
\[ HD = BH \cdot \tan C, \quad HE = AH \cdot \tan A, \quad HF = CH \cdot \tan B \]
Using the relationships:
\[ HD = BH \cdot \cot C = (2R \cos B) \cdot \cot C = 2R \cos B \cdot \frac{\cos C}{\sin C \]
Thus, we can express:
\[ HD = \frac{2R \cos B \cos C}{\sin C} \]
Similarly, we can find:
\[ HE = \frac{2R \cos A \cos C}{\sin A}, \quad HF = \frac{2R \cos A \cos B}{\sin B} \]

6. **Calculating the Product**:
Now we can calculate the product \( HD \cdot HE \cdot HF \):
\[ HD \cdot HE \cdot HF = \left(\frac{2R \cos B \cos C}{\sin C}\right) \cdot \left(\frac{2R \cos A \cos C}{\sin A}\right) \cdot \left(\frac{2R \cos A \cos B}{\sin B}\right) \]
This simplifies to:
\[ = \frac{(2R)^3 \cos^2 A \cos^2 B \cos^2 C}{\sin A \sin B \sin C} \]

7. **Using the Known Values**:
From equation (1), we have:
\[ 8R^3 \cos A \cos B \cos C = 3 \implies R^3 \cos A \cos B \cos C = \frac{3}{8} \]
Thus, substituting this into the product gives:
\[ HD \cdot HE \cdot HF = \frac{3}{8} \cdot \frac{8}{\sin A \sin B \sin C} = \frac{3}{\sin A \sin B \sin C} \]

8. **Final Result**:
The final value of \( HD \cdot HE \cdot HF \) is:
\[ HD \cdot HE \cdot HF = \frac{9}{8} R^3 \]

### Conclusion:
The value of \( HD \cdot HE \cdot HF \) is \( \frac{9}{8} R^3 \).