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 `(cos A. cos B . cos C)/(cos^(2)A + cos^(2)B + cos^(2)C)` is

Options

A. `(3)/(14R)`
B. `(3)/(7R)`
C. `(7)/(3R)`
D. `(14)/(3R)`

Explanation

To solve the problem, we need to find the value of \(\frac{\cos A \cdot \cos B \cdot \cos C}{\cos^2 A + \cos^2 B + \cos^2 C}\) given the conditions about the orthocenter and the lengths from the orthocenter to the vertices of triangle \(ABC\).

### Step 1: Define the lengths from the orthocenter
Let:
- \(AH = 2R \cos A\)
- \(BH = 2R \cos B\)
- \(CH = 2R \cos C\)

### Step 2: Use the given product of lengths
We are given that:
\[ AH \cdot BH \cdot CH = 3 \]
Substituting the expressions we defined:
\[ (2R \cos A)(2R \cos B)(2R \cos C) = 3 \]
This simplifies to:
\[ 8R^3 \cos A \cos B \cos C = 3 \]
Thus, we can express \(\cos A \cos B \cos C\) as:
\[ \cos A \cos B \cos C = \frac{3}{8R^3} \quad \text{(Equation 1)} \]

### Step 3: Use the given sum of squares of lengths
We are also given:
\[ AH^2 + BH^2 + CH^2 = 7 \]
Substituting the expressions:
\[ (2R \cos A)^2 + (2R \cos B)^2 + (2R \cos C)^2 = 7 \]
This simplifies to:
\[ 4R^2 (\cos^2 A + \cos^2 B + \cos^2 C) = 7 \]
Thus, we can express \(\cos^2 A + \cos^2 B + \cos^2 C\) as:
\[ \cos^2 A + \cos^2 B + \cos^2 C = \frac{7}{4R^2} \quad \text{(Equation 2)} \]

### Step 4: Substitute into the main expression
Now we substitute Equations 1 and 2 into the expression we want to evaluate:
\[ \frac{\cos A \cos B \cos C}{\cos^2 A + \cos^2 B + \cos^2 C} = \frac{\frac{3}{8R^3}}{\frac{7}{4R^2}} \]
This simplifies to:
\[ = \frac{3}{8R^3} \cdot \frac{4R^2}{7} = \frac{3 \cdot 4}{8 \cdot 7} \cdot \frac{1}{R} = \frac{12}{56R} = \frac{3}{14R} \]

### Final Answer
Thus, the value of \(\frac{\cos A \cdot \cos B \cdot \cos C}{\cos^2 A + \cos^2 B + \cos^2 C}\) is:
\[ \frac{3}{14R} \]