We begin with the system:

$$ \begin{aligned} x+y+2z &= 6, \\ 2x+3y+az &= a+1, \\ -x-3y+bz &= 2b. \end{aligned} $$

Step 1. Solve the first equation for $$x$$:

$$ x = 6 - y - 2z. $$

Step 2. Substitute $$x = 6-y-2z$$ into the second equation:

$$ 2(6-y-2z) + 3y + az = a+1. $$

Expanding and simplifying:

$$ 12 - 2y - 4z + 3y + az = a+1 \quad \Longrightarrow \quad y + (a-4)z = a - 11. $$

Call this Equation (I).

Step 3. Substitute $$x = 6-y-2z$$ into the third equation:

$$ -(6-y-2z) - 3y + bz = 2b. $$

Expanding and simplifying:

$$ -6 + y + 2z - 3y + bz = 2b \quad \Longrightarrow \quad -2y + (b+2)z = 2b + 6. $$

Call this Equation (II).

Step 4. For the system to have infinitely many solutions, the two equations in $$y$$ and $$z$$ must be dependent—that is, one must be a constant multiple of the other. Assume there exists a constant $$k$$ such that

$$ -2 = k \cdot 1 \quad \Longrightarrow \quad k = -2. $$

Apply this to the coefficient of $$z$$ and the constant term.

For the $$z$$-coefficient in Equations (I) and (II):

$$ b+2 = k(a-4) = -2(a-4) = -2a+8. $$

Thus,

$$ b = -2a+6. $$

For the constant term:

$$ 2b+6 = k(a-11) = -2(a-11) = -2a + 22. $$

Substitute $$b = -2a+6$$ into this equation:

$$ 2(-2a+6) + 6 = -2a + 22 \quad \Longrightarrow \quad -4a + 12 + 6 = -2a + 22. $$

Simplify:

$$ -4a + 18 = -2a + 22. $$

Solve for $$a$$:

$$ -4a + 18 + 4a = -2a + 22 + 4a \quad \Longrightarrow \quad 18 = 2a + 22, $$

$$ 2a = 18 - 22 = -4 \quad \Longrightarrow \quad a = -2. $$

Substitute $$a = -2$$ into $$b = -2a+6$$:

$$ b = -2(-2) + 6 = 4 + 6 = 10. $$

Step 5. We now compute

$$ 7a + 3b = 7(-2) + 3(10) = -14 + 30 = 16. $$

Thus, the value of $$7a+3b$$ is

$$ \boxed{16}. $$