If the lines $$2x + 3y + 1 = 0$$ and $$3x - y - 4 = 0$$ lie along diameters of a circle of circumference $$10\pi$$, then the equation of the circle is

Since the given lines lie along diameters of the circle, their point of intersection is the centre of the circle.

Solving $$2x+3y+1=0$$ and $$3x-y-4=0,$$

from the second equation,

$$y=3x-4.$$

Substituting into the first equation,

$$2x+3(3x-4)+1=0$$

$$11x-11=0$$

$$x=1.$$

Therefore,

$$y=3(1)-4=-1.$$

Hence the centre of the circle is $$C(1,-1).$$

The circumference is $$10\pi.$$

Using $$2\pi r=10\pi,$$

we get $$r=5.$$

Therefore,

$$r^2=25.$$

The equation of the circle is $$(x-1)^2+(y+1)^2=25.$$

Expanding,

$$x^2-2x+1+y^2+2y+1=25.$$

$$x^2+y^2-2x+2y-23=0.$$

Hence the equation of the circle is $$\boxed{x^2+y^2-2x+2y-23=0}.$$