5. Given the line $l_{1}: y=a x+3 a+2$ and $l_{2}: y$ $=-3 x+3$ intersect at a point in the first quadrant. Then the range of values for $a$ is $(\quad)$.
(A) $\left(-\frac{1}{2}, \frac{1}{3}\right)$
(B) $\left(-\infty, \frac{1}{3}\right)$
(C) $\left(-3, \frac{1}{3}\right)$
(D) $\left(-\frac{1}{2},+\infty\right)$
5. (A).

As shown in Figure 6, $l_{1}$ is a family of lines passing through the fixed point $M(-3,2)$; $l_{2}$ intersects the $x$-axis at $A(1,0)$ and the $y$-axis at $B(0,3)$. For the intersection point of the two lines to be in the first quadrant, it must and only must lie on the line segment $AB$. First, the points on the line segment $AB$ can be represented as
$$
\left\{\begin{array}{l}
x=\frac{1}{1+\lambda}, \\
y=\frac{3 \lambda}{1+\lambda} .
\end{array}(\lambda>0)\right.
$$

Substituting into $l_{1}$ gives
$$
\frac{3 \lambda}{1+\lambda}=\frac{a}{1+\lambda}+(3 a+2),
$$

which simplifies to $\lambda=\frac{2(2 a+1)}{1-3 a}>0$.
Solving this yields $-\frac{1}{2}<a<\frac{1}{3}$.
Analysis: The geometric meaning of the problem is that the slope of the line connecting the intersection point with $M$ lies between $k_{A M}=-\frac{1}{2}$ and $k_{B M}=\frac{1}{3}$.