Let the line $$2x + 3y - k = 0, k > 0$$, intersect the $$x$$-axis and $$y$$-axis at the points $$A$$ and $$B$$, respectively. If the equation of the circle having the line segment $$AB$$ as a diameter is $$x^2 + y^2 - 3x - 2y = 0$$ and the length of the latus rectum of the ellipse $$x^2 + 9y^2 = k^2$$ is $$\frac{m}{n}$$, where $$m$$ and $$n$$ are coprime, then $$2m + n$$ is equal to

We need to find $$k$$ from the circle equation, then use it to find the latus rectum of the ellipse, and compute $$2m + n$$. The line is $$2x + 3y - k = 0$$ with $$k > 0$$. Its x-intercept is found by setting $$y = 0$$, which gives $$2x = k \implies x = \frac{k}{2}$$ so point $$A = \left(\frac{k}{2}, 0\right)$$, and its y-intercept by setting $$x = 0$$, which gives $$3y = k \implies y = \frac{k}{3}$$ so $$B = \left(0, \frac{k}{3}\right)$$.

The equation of the circle with AB as diameter follows from the general form for endpoints $$(x_1, y_1)$$ and $$(x_2, y_2)$$: $$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$ which becomes $$\left(x - \frac{k}{2}\right)(x - 0) + \left(y - 0\right)\left(y - \frac{k}{3}\right) = 0$$ and simplifies to $$x^2 - \frac{k}{2}x + y^2 - \frac{k}{3}y = 0$$.

Comparing this with the given circle equation $$x^2 + y^2 - 3x - 2y = 0$$ shows that $$\frac{k}{2} = 3$$ and $$\frac{k}{3} = 2$$, both yielding $$k = 6$$.

Substituting $$k = 6$$ into the ellipse equation gives $$x^2 + 9y^2 = 36$$, which may be written as $$\frac{x^2}{36} + \frac{y^2}{4} = 1$$. Here $$a^2 = 36$$ and $$b^2 = 4$$, so $$a = 6$$ and $$b = 2$$. The length of the latus rectum is given by $$\ell = \frac{2b^2}{a} = \frac{2 \times 4}{6} = \frac{8}{6} = \frac{4}{3}$$. Writing the latus rectum as $$\frac{m}{n} = \frac{4}{3}$$ with coprime integers $$m = 4$$ and $$n = 3$$, one finds $$2m + n = 2(4) + 3 = 11$$.

The correct answer is Option (1): 11.