8. Let the line $l_{1}: 3 x-2 y=1$, passing through point $A(-1,-2)$, line $l_{2}: y=1$ intersects $l_{1}$ at point $B$, and line $l_{3}$ has a positive slope, passing through point $A$, and intersects line $l_{2}$ at point $C$. If $S_{\triangle A B C}=3$, then the slope of line $l_{3}$ is ( ).
(A) $\frac{2}{3}$
(B) $\frac{3}{4}$
(C) 1
(D) $\frac{4}{3}$
(E) $\frac{3}{2}$
8. B.

Since $l_{2}: y=1$, point $A(-1,-2)$, and $S_{\triangle A B C}=3$, therefore, $B C=2$.
From $\left\{\begin{array}{l}3 x-2 y=1, \\ y=1,\end{array}\right.$ we get point $B(1,1)$.
Thus, point $C(3,1)$ or $C(-1,1)$.
Since the slope of $l_{3}$ is positive, point $C(3,1)$.
Therefore, the slope of $l_{3}$ is $k=\frac{1+2}{3+1}=\frac{3}{4}$.