STATEMENT-1 : The area of equilateral triangle inscribed in the circle `x^(2) + y^(2) + 6x + 8y + 24 = 0` is `(3sqrt(3))/(4)`.
STATEMENT-2 : The abscissae of two points A and B are the roots of the equation `x^(2) + 3ax + b^(2) =0` and their ordinates are the roots of `x^(2) + 3bx + a^(2) =0` Then the equation of the circle with AB as diameter is `x^(2) + y^(2) + 3ax + 3by + a^(2) + b^(2) = 0`.
STATEMENT-3 : If the circle `x^(2) + y^(2) + 2gx + 2fx +c =0` always passes through exactly three quadrants not passing through origin then `c gt 0`.

Options

A. T F T
B. T T T
C. F F F
D. F F T

Answer

To solve the question step by step, we will analyze each statement one by one.

### Statement 1:
**The area of the equilateral triangle inscribed in the circle \( x^2 + y^2 + 6x + 8y + 24 = 0 \) is \( \frac{3\sqrt{3}}{4} \).**

1. **Convert the circle equation to standard form:**
\[ x^2 + y^2 + 6x + 8y + 24 = 0 \]
Completing the square for \( x \) and \( y \):
\[ (x^2 + 6x) + (y^2 + 8y) + 24 = 0 \]
\[ (x + 3)^2 - 9 + (y + 4)^2 - 16 + 24 = 0 \]
\[ (x + 3)^2 + (y + 4)^2 = 1 \]
Thus, the center of the circle is \( (-3, -4) \) and the radius \( r = 1 \).

2. **Area of the equilateral triangle inscribed in the circle:**
The area \( A \) of an equilateral triangle inscribed in a circle of radius \( r \) is given by:
\[ A = \frac{3\sqrt{3}}{4} r^2 \]
Substituting \( r = 1 \):
\[ A = \frac{3\sqrt{3}}{4} \cdot 1^2 = \frac{3\sqrt{3}}{4} \]
Therefore, Statement 1 is **True**.

### Statement 2:
**The abscissae of two points A and B are the roots of the equation \( x^2 + 3ax + b^2 = 0 \) and their ordinates are the roots of \( x^2 + 3bx + a^2 = 0 \). The equation of the circle with AB as diameter is \( x^2 + y^2 + 3ax + 3by + a^2 + b^2 = 0 \).**

1. **Let the roots of the first equation be \( x_1 \) and \( x_2 \):**
- Sum of roots \( x_1 + x_2 = -3a \)
- Product of roots \( x_1 x_2 = b^2 \)

2. **Let the roots of the second equation be \( y_1 \) and \( y_2 \):**
- Sum of roots \( y_1 + y_2 = -3b \)
- Product of roots \( y_1 y_2 = a^2 \)

3. **Equation of the circle with AB as diameter:**
The equation of a circle with diameter endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \) is:
\[ \left(x - \frac{x_1 + x_2}{2}\right)^2 + \left(y - \frac{y_1 + y_2}{2}\right)^2 = \left(\frac{x_2 - x_1}{2}\right)^2 + \left(\frac{y_2 - y_1}{2}\right)^2 \]
Expanding this gives:
\[ x^2 + y^2 + 3ax + 3by + a^2 + b^2 = 0 \]
Therefore, Statement 2 is **True**.

### Statement 3:
**If the circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \) always passes through exactly three quadrants not passing through the origin, then \( c > 0 \).**

1. **Analyzing the condition:**
- The circle passes through three quadrants, which implies it must be positioned such that it does not intersect the x-axis and y-axis in the fourth quadrant.
- The center of the circle is \( (-g, -f) \).

2. **Condition for the circle to not pass through the origin:**
- Substituting \( (0, 0) \) into the circle's equation gives \( c > 0 \).

3. **Conclusion:**
- If \( c > 0 \), then the circle will not pass through the origin and can be positioned to pass through three quadrants.
Therefore, Statement 3 is **True**.

### Final Conclusion:
All three statements are true.