jmc

We find the equation of $l_1$ first. The slope of $l_1$ is $$\frac{14-8}{7-5}=3.$$Since $l_1$ passes through $5,8$, we can set $8=3(5)+t$ where $(0,t)$ is the $y$-intercept. Subtracting 15 from both sides, we find the $t=-7$, and so the $y$-intercept is $(0,-7)$. Thus, $l_1$ has equation $y=3x-7$.

To find the equation of the original line, we begin with line $l_1$ and shift it three units to the right and two units up. Recall that to shift the graph of an equation three units to the right we replace $x$ with $x-3$. Similarly, to shift two units up we replace $y$ with $y-2$. Making these substitutions, we obtain $y-2=3(x-3)+7$, which is equivalent to $y=3x-14$. Thus, $a-b=3+14=\boxed{17}$.