The positive real numbers $$a, b, c$$ satisfy: $$\frac{a}{2b+1}+\frac{2b}{3c+1}+\frac{3c}{a+1}=1$$ and $$\frac{1}{a+1}+\frac{1}{2b+1}+\frac{1}{3c+1}=2$$. What is the value of $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$?

Let us rewrite the two given relations in a more symmetric way.

Put $$A = a+1,\; B = 2b+1,\; C = 3c+1$$.
Since $$a,b,c \gt 0$$, we have $$A,B,C \gt 1$$.

Then

$$a = A-1,\qquad 2b = B-1 \; \Longrightarrow\; b = \frac{B-1}{2},\qquad 3c = C-1 \; \Longrightarrow\; c = \frac{C-1}{3}.$$

Substituting in the two equations:

1. $$\frac{a}{2b+1}+\frac{2b}{3c+1}+\frac{3c}{a+1}=1$$ becomes
$$\frac{A-1}{B}+\frac{B-1}{C}+\frac{C-1}{A}=1.$$

2. $$\frac{1}{a+1}+\frac{1}{2b+1}+\frac{1}{3c+1}=2$$ becomes
$$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=2.$$

Introduce the reciprocals

$$x=\frac1A,\; y=\frac1B,\; z=\frac1C\qquad(0\lt x,y,z\lt 1).$$

With these, the second relation is simply

$$x+y+z=2 \; \; -(1).$$

The first relation changes to

$$\left(\frac1x-1\right)\!y+\left(\frac1y-1\right)\!z+\left(\frac1z-1\right)\!x =1,$$ which simplifies to $$\frac{y}{x}+\frac{z}{y}+\frac{x}{z}=3\; \; -(2).$$

Observe that for any positive numbers, the arithmetic-geometric mean inequality gives

$$\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\;\ge\;3,$$ with equality only when $$\frac{y}{x}=\frac{z}{y}=\frac{x}{z}=1.$$

Because the left-hand side is exactly $$3$$ by (2), equality must occur, hence

$$\frac{y}{x}=1,\;\; \frac{z}{y}=1,\;\; \frac{x}{z}=1 \;\Longrightarrow\; x=y=z.$$

Using (1): $$3x=2\;\Longrightarrow\; x=y=z=\frac23.$$

This immediately gives

$$A=\frac1x=\frac32,\quad B=\frac1y=\frac32,\quad C=\frac1z=\frac32.$$

Recover $$a,b,c$$:

$$a=A-1=\frac32-1=\frac12,$$ $$b=\frac{B-1}{2}=\frac{\frac32-1}{2}=\frac14,$$ $$c=\frac{C-1}{3}=\frac{\frac32-1}{3}=\frac16.$$

Finally, compute the required sum:

$$\frac1a+\frac1b+\frac1c =\frac1{\frac12}+\frac1{\frac14}+\frac1{\frac16} =2+4+6=12.$$

Hence the value of $$\displaystyle\frac1a+\frac1b+\frac1c$$ is 12.