If the area of an equilateral triangle inscribed in the circle $$x^2 + y^2 + 10x + 12y + c = 0$$ is $$27\sqrt{3}$$ sq. units, then $$c$$ is equal to:

We start with the given circle

$$x^{2}+y^{2}+10x+12y+c=0.$$

To recognise its centre and radius, we complete the squares. Write the quadratic terms together:

$$\bigl(x^{2}+10x\bigr)+\bigl(y^{2}+12y\bigr)= -\,c.$$

Now add and subtract the necessary constants inside each bracket. The coefficient of $$x$$ is $$10$$, so half of it is $$5$$ and its square is $$25$$. For $$y$$, the coefficient is $$12$$, half is $$6$$ and its square is $$36$$. Inserting these we get

$$\bigl(x^{2}+10x+25-25\bigr)+\bigl(y^{2}+12y+36-36\bigr)= -\,c.$$

Group the perfect squares and the constants separately:

$$(x+5)^{2}+(y+6)^{2}-25-36= -\,c.$$

Combine the constants $$-25-36=-61$$ and move them to the right side:

$$(x+5)^{2}+(y+6)^{2}=61-c.$$

Hence the circle is centred at $$(-5,-6)$$ and its radius is

$$R=\sqrt{61-c}.$$

Next, an equilateral triangle is inscribed in this circle. For an equilateral triangle the standard relation between the side length $$a$$ and the circum-radius $$R$$ is

$$R=\frac{a}{\sqrt{3}}$$ (circumradius formula for an equilateral triangle).

The area formula for an equilateral triangle is

$$\text{Area}=\frac{\sqrt{3}}{4}\,a^{2}.$$

We are told that the area equals $$27\sqrt{3}$$, so substituting this value gives

$$\frac{\sqrt{3}}{4}\,a^{2}=27\sqrt{3}.$$

Divide both sides by $$\sqrt{3}$$:

$$\frac{1}{4}\,a^{2}=27.$$

Multiply by $$4$$:

$$a^{2}=108.$$

Take the square root (the side length is positive):

$$a=\sqrt{108}=6\sqrt{3}.$$

Now substitute this $$a$$ into the circumradius formula:

$$R=\frac{a}{\sqrt{3}}=\frac{6\sqrt{3}}{\sqrt{3}}=6.$$

Thus the radius of the given circle must be

$$R=6.$$

But from the circle’s equation we already have

$$R=\sqrt{61-c}.$$

Equate the two expressions for $$R$$:

$$\sqrt{61-c}=6.$$

Square both sides to remove the square root:

$$61-c=36.$$

Solve for $$c$$ by moving terms:

$$c=61-36=25.$$

Hence, the correct answer is Option A.