The equation of the circle passing through the point $$(1, 2)$$ and through the points of intersection of $$x^2 + y^2 - 4x - 6y - 21 = 0$$ and $$3x + 4y + 5 = 0$$ is given by

The points of intersection of the given circle $$x^{2}+y^{2}-4x-6y-21=0$$ and the straight line $$3x+4y+5=0$$ can be used to generate a family of circles.

General form of the family of circles passing through the intersection of a circle $$S(x,y)=0$$ and a line $$L(x,y)=0$$ is
$$S(x,y)+\lambda\,L(x,y)=0,$$
where $$\lambda$$ is a real parameter.

Here,
$$S(x,y)=x^{2}+y^{2}-4x-6y-21,$$
$$L(x,y)=3x+4y+5.$$

Therefore the required family is
$$x^{2}+y^{2}-4x-6y-21+\lambda(3x+4y+5)=0.$$

Simplify the coefficients:
$$x^{2}+y^{2}+(-4+3\lambda)\,x+(-6+4\lambda)\,y+(-21+5\lambda)=0.$$

The required circle must also pass through the point $$(1,2)$$. Substitute $$x=1,\;y=2$$:

$$\begin{aligned} 1^{2}+2^{2}&+(-4+3\lambda)(1)+(-6+4\lambda)(2)+(-21+5\lambda)=0\\ 5&+(-4+3\lambda)+(-12+8\lambda)+(-21+5\lambda)=0\\ \; &\Longrightarrow -32+16\lambda=0\\ \; &\Longrightarrow \lambda=2. \end{aligned}$$

Put $$\lambda=2$$ back into the family’s coefficients:

Coefficient of $$x$$: $$-4+3(2)=2,$$
Coefficient of $$y$$: $$-6+4(2)=2,$$
Constant term: $$-21+5(2)=-11.$$

Hence the required circle is
$$x^{2}+y^{2}+2x+2y-11=0.$$

Comparing with the options, this matches Option D.

Option D which is: $$x^2 + y^2 + 2x + 2y - 11 = 0$$