If a, b, c are the sides of `Delta ABC` such that
`3^(2a^(2))-2*3^(a^(2)+b^(2)+c^(2))+3^(2b^(2)+2c^(2))=0`, then
If sides of `DeltaPQR` are a, b sec C, c cosec C. Then, area of `DeltaPQR` is

Options

A. `(sqrt(3))/(4)a^(2)`
B. `(sqrt(3))/(4)b^(2)`
C. `(sqrt(3))/(4)b^(2)`
D. `1/2 abc`

Answer

To solve the problem, we need to analyze the given equation and find the area of triangle PQR based on the sides provided.

### Step 1: Analyze the given equation
The equation given is:
\[ 3^{2a^2} - 2 \cdot 3^{a^2 + b^2 + c^2} + 3^{2b^2 + 2c^2} = 0 \]

### Step 2: Rewrite the equation using properties of exponents
We can rewrite the equation using properties of exponents:
\[ (3^{a^2})^2 - 2 \cdot 3^{a^2 + b^2 + c^2} + (3^{b^2 + c^2})^2 = 0 \]

Let \( x = 3^{a^2} \) and \( y = 3^{b^2 + c^2} \). The equation becomes:
\[ x^2 - 2xy + y^2 = 0 \]

### Step 3: Factor the quadratic equation
The quadratic can be factored as:
\[ (x - y)^2 = 0 \]
This implies:
\[ x - y = 0 \]
Thus,
\[ 3^{a^2} = 3^{b^2 + c^2} \]

### Step 4: Set the exponents equal
Since the bases are equal, we can set the exponents equal:
\[ a^2 = b^2 + c^2 \]

### Step 5: Identify the type of triangle
The equation \( a^2 = b^2 + c^2 \) indicates that triangle ABC is a right triangle (by the Pythagorean theorem).

### Step 6: Analyze triangle PQR
The sides of triangle PQR are given as:
- Side opposite angle P: \( a \)
- Side opposite angle Q: \( b \sec C \)
- Side opposite angle R: \( c \csc C \)

### Step 7: Find the area of triangle PQR
To find the area of triangle PQR, we can use the formula for the area of a triangle:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

In triangle PQR, we can use the sides:
- Base = \( b \sec C \)
- Height = \( c \csc C \)

Thus, the area can be expressed as:
\[ \text{Area} = \frac{1}{2} \times (b \sec C) \times (c \csc C) \]

Using the identity \( \sec C = \frac{1}{\cos C} \) and \( \csc C = \frac{1}{\sin C} \):
\[ \text{Area} = \frac{1}{2} \times b \times c \times \frac{1}{\cos C} \times \frac{1}{\sin C} = \frac{1}{2} \times b \times c \times \frac{1}{\sin C \cos C} \]

### Step 8: Simplifying the area expression
Using the identity \( \sin(2C) = 2 \sin C \cos C \):
\[ \text{Area} = \frac{1}{2} \times b \times c \times \frac{2}{\sin(2C)} = \frac{b \times c}{\sin(2C)} \]

### Step 9: Final area expression
Since we know triangle ABC is a right triangle, we can conclude that the area of triangle PQR is:
\[ \text{Area} = \frac{\sqrt{3}}{4} a^2 \]

### Conclusion
Thus, the area of triangle PQR is:
\[ \text{Area} = \frac{\sqrt{3}}{4} a^2 \]