smc

Line $l_1$ has the equation $y=3x/2-1/2$ when rearranged. Substituting $1$ for $y$, we find that line $l_2$ will meet this line at point $(1,1)$, which is point $B$. We call $\overline{BC}$ the base and the altitude from $A$ to the line connecting $B$ and $C$, $y=-1$, the height. The altitude has length $|-2-1|=3$, and the area of $△ABC=3$. Since $A=bh/2$, $b=2$. Because $l_3$ has positive slope, it will meet $l_2$ to the right of $B$, and the point $2$ to the right of $B$ is $(3,1)$. $l_3$ passes through $(-1,-2)$ and $(3,1)$, and thus has slope $\frac{|1-(-2)|}{|3-(-1)|}=$ $\boxed{\frac{3}{4}}$.