The equation of the circle described on the chord $$3x + y + 5 = 0$$ of the circle $$x^2 + y^2 = 16$$ as the diameter is:

The given circle is $$x^2 + y^2 = 16$$, which has center (0,0) and radius 4. The chord $$3x + y + 5 = 0$$ is given, and we need to find the circle that has this chord as its diameter.

First, we find the points where the chord intersects the given circle. These points will be the endpoints of the diameter for the new circle. Substitute $$y = -3x - 5$$ from the chord equation into the circle equation:

$$x^2 + (-3x - 5)^2 = 16$$

Expand the expression:

$$x^2 + (9x^2 + 30x + 25) = 16$$

Combine like terms:

$$10x^2 + 30x + 25 = 16$$

Bring all terms to one side:

$$10x^2 + 30x + 25 - 16 = 0$$

$$10x^2 + 30x + 9 = 0$$

Solve this quadratic equation using the quadratic formula, where $$a = 10$$, $$b = 30$$, $$c = 9$$:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Calculate the discriminant:

$$D = b^2 - 4ac = 30^2 - 4 \times 10 \times 9 = 900 - 360 = 540$$

So,

$$x = \frac{-30 \pm \sqrt{540}}{20}$$

Simplify $$\sqrt{540}$$:

$$\sqrt{540} = \sqrt{36 \times 15} = 6\sqrt{15}$$

Thus,

$$x = \frac{-30 \pm 6\sqrt{15}}{20} = \frac{-15 \pm 3\sqrt{15}}{10}$$

So the x-coordinates are:

$$x_1 = \frac{-15 + 3\sqrt{15}}{10}, \quad x_2 = \frac{-15 - 3\sqrt{15}}{10}$$

Now find the corresponding y-coordinates using $$y = -3x - 5$$:

For $$x_1$$:

$$y_1 = -3 \left( \frac{-15 + 3\sqrt{15}}{10} \right) - 5 = \frac{45 - 9\sqrt{15}}{10} - 5 = \frac{45 - 9\sqrt{15}}{10} - \frac{50}{10} = \frac{-5 - 9\sqrt{15}}{10}$$

For $$x_2$$:

$$y_2 = -3 \left( \frac{-15 - 3\sqrt{15}}{10} \right) - 5 = \frac{45 + 9\sqrt{15}}{10} - 5 = \frac{45 + 9\sqrt{15}}{10} - \frac{50}{10} = \frac{-5 + 9\sqrt{15}}{10}$$

The endpoints of the diameter are:

$$A\left( \frac{-15 + 3\sqrt{15}}{10}, \frac{-5 - 9\sqrt{15}}{10} \right), \quad B\left( \frac{-15 - 3\sqrt{15}}{10}, \frac{-5 + 9\sqrt{15}}{10} \right)$$

The center of the new circle is the midpoint of AB. Find the midpoint:

Midpoint x-coordinate:

$$\frac{x_1 + x_2}{2} = \frac{ \frac{-15 + 3\sqrt{15}}{10} + \frac{-15 - 3\sqrt{15}}{10} }{2} = \frac{ \frac{-30}{10} }{2} = \frac{-3}{2} = -\frac{3}{2}$$

Midpoint y-coordinate:

$$\frac{y_1 + y_2}{2} = \frac{ \frac{-5 - 9\sqrt{15}}{10} + \frac{-5 + 9\sqrt{15}}{10} }{2} = \frac{ \frac{-10}{10} }{2} = \frac{-1}{2} = -\frac{1}{2}$$

So the center is $$\left( -\frac{3}{2}, -\frac{1}{2} \right)$$.

Now find the radius by computing the distance from the center to point A. Write the center as $$\left( -\frac{15}{10}, -\frac{5}{10} \right)$$ for ease:

Difference in x-coordinates:

$$\frac{-15 + 3\sqrt{15}}{10} - \left( -\frac{15}{10} \right) = \frac{-15 + 3\sqrt{15} + 15}{10} = \frac{3\sqrt{15}}{10}$$

Difference in y-coordinates:

$$\frac{-5 - 9\sqrt{15}}{10} - \left( -\frac{5}{10} \right) = \frac{-5 - 9\sqrt{15} + 5}{10} = \frac{-9\sqrt{15}}{10}$$

Distance squared:

$$\left( \frac{3\sqrt{15}}{10} \right)^2 + \left( \frac{-9\sqrt{15}}{10} \right)^2 = \frac{9 \times 15}{100} + \frac{81 \times 15}{100} = \frac{135}{100} + \frac{1215}{100} = \frac{1350}{100} = \frac{27}{2}$$

So the radius squared is $$\frac{27}{2}$$. The equation of the circle is:

$$\left( x + \frac{3}{2} \right)^2 + \left( y + \frac{1}{2} \right)^2 = \frac{27}{2}$$

Expand the left side:

$$x^2 + 2 \times x \times \frac{3}{2} + \left( \frac{3}{2} \right)^2 + y^2 + 2 \times y \times \frac{1}{2} + \left( \frac{1}{2} \right)^2 = x^2 + 3x + \frac{9}{4} + y^2 + y + \frac{1}{4} = x^2 + y^2 + 3x + y + \frac{10}{4} = x^2 + y^2 + 3x + y + \frac{5}{2}$$

Set equal to $$\frac{27}{2}$$:

$$x^2 + y^2 + 3x + y + \frac{5}{2} = \frac{27}{2}$$

Bring all terms to the left:

$$x^2 + y^2 + 3x + y + \frac{5}{2} - \frac{27}{2} = 0$$

$$x^2 + y^2 + 3x + y - \frac{22}{2} = 0$$

$$x^2 + y^2 + 3x + y - 11 = 0$$

Comparing with the options, this matches option C.

Hence, the correct answer is Option C.