5. The triangle with vertices $A(-3,0)$, $B(0,-3)$, and $C\left(\frac{15}{7}, \frac{24}{7}\right)$ has no common points with the circle $x^{2}+y^{2}=R^{2}(R>0)$. Then the range of the radius $R$ is ( ).
(A) $\left(\frac{3 \sqrt{2}}{2}, 3\right)$
(B) $\left(0, \frac{3 \sqrt{2}}{2}\right) \cup(3,+\infty)$
(C) $\left(\frac{3 \sqrt{10}}{10}, \frac{3 \sqrt{89}}{7}\right)$
(D) $\left(0, \frac{3 \sqrt{10}}{10}\right) \cup\left(\frac{3 \sqrt{89}}{7},+\infty\right)$
5.D.

As shown in Figure 3, from the given information we have
$$
\begin{array}{l}
|O A|=3, \\
|O B|=3, \\
|O C|=\frac{3 \sqrt{89}}{7} .
\end{array}
$$

When $R > |O C|$, the triangle and the circle have no intersection points.

The equations of the lines $A B$, $A C$, and $B C$ are respectively
$$
y=-x-3, y=\frac{2}{3}(x+3), y=3(x-1),
$$

The distances from the origin (the center of the circle) to these three sides are respectively
$$
d_{A B}=\frac{3 \sqrt{2}}{2}, d_{A C}=\frac{6 \sqrt{13}}{13}, d_{B C}=\frac{3 \sqrt{10}}{10} .
$$

When $R$ is less than the smallest of these values, i.e., $0<R<\frac{3 \sqrt{10}}{10}$, the circle and the triangle also have no common points. Therefore, the range of $R$ is
$$
\left(0, \frac{3 \sqrt{10}}{10}\right) \cup\left(\frac{3 \sqrt{89}}{7},+\infty\right) .
$$