5. Given the quadratic inequality $a x^{2}+b x+c \geqslant 0$ has the solution set $-2 \leqslant x \leqslant 3$. Then the solution set of the inequality $c x^{2}+b x+a \leqslant 0$ is ( ).
(A) $x \geqslant 3$ or $x \leqslant-2$
(B) $x \leqslant-\frac{1}{2}$ or $x \geqslant \frac{1}{3}$
(C) $-\frac{1}{2} \leqslant x \leqslant \frac{1}{3}$
(D) $-2 \leqslant x \leqslant 3$
5. (B).

From the problem, we have $a<0$, and
$$
\begin{array}{l}
a x^{2}+b x+c \\
=a(x+2)(x-3)=a x^{2}-a x-6 a,
\end{array}
$$

then $c x^{2}+b x+a$
$$
\begin{aligned}
= & -6 a x^{2}-a x+a \\
& =-a(3 x-1)(2 x+1) \leqslant 0
\end{aligned}
$$

Thus, the solution set is $x \leqslant-\frac{1}{2}$ or $x \geqslant \frac{1}{3}$.