Step 1: Understand what a bijection requires.
A function is a bijection when it is both one-one and onto. For $f(x) = \frac{x+3}{x-2}$, the domain excludes the value that makes the denominator zero, and the codomain excludes the value that $f(x)$ can never equal.
Step 2: Find $l$ (the excluded domain value).
The denominator is $x - 2$. It equals zero when $x = 2$. So the domain is $\mathbb{R} - \{2\}$, meaning $l = 2$.
Step 3: Find $m$ (the excluded range value).
Set $y = \frac{x+3}{x-2}$ and solve for $x$: \[ y(x-2) = x+3 \implies xy - 2y = x + 3 \implies x(y-1) = 2y+3 \implies x = \frac{2y+3}{y-1} \] This is undefined when $y = 1$, so $f(x)$ can never equal $1$. Hence $m = 1$ and the range is $\mathbb{R} - \{1\}$.
Step 4: Confirm that this is indeed a bijection.
With domain $\mathbb{R} - \{2\}$ and codomain $\mathbb{R} - \{1\}$, every value $y \neq 1$ is achieved by exactly one $x \neq 2$, so it is a bijection.
Step 5: Compute $3l + 2m$.
\[ 3l + 2m = 3(2) + 2(1) = 6 + 2 = 8 \]
Step 6: State the answer.
\[ \boxed{8} \]