5. As shown in Figure 2, point $O$ is inside $\triangle A B C$, and points $P, Q, R$ are on sides $A B, B C, C A$ respectively, with $O P / / B C$, $A B, O P=O Q=O R$ $=x$. Let $B C=a, C A$ $=b, A B=c$. Then $x$ equals ( ).
(A) $\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$
(B) $\frac{1}{9}(a+b+c)$
(C) $\frac{1}{3} \sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}}$
(D) $\frac{1}{3} \sqrt{\frac{a b+b c+c a}{3}}$
5. A.

As shown in Figure 5, construct $P K \parallel O Q, Q L \parallel O R, R M \parallel O P$, intersecting $B C, C A, A B$ at points $K, L, M$ respectively. It is easy to see that
$$
\begin{array}{l}
P K=K Q=Q L \\
=O R=x .
\end{array}
$$

Let $B K=y, Q C=z$, then $x+y+z=B C=a$.
From $\frac{P K}{A C}=\frac{B K}{B C}, \frac{Q L}{A B}=\frac{Q C}{B C}$ we have
$$
\frac{x}{b}=\frac{y}{a}, \frac{x}{c}=\frac{z}{a} \text {. }
$$

Substituting into equation (1) gives
$$
\frac{a}{a} x+\frac{a}{b} x+\frac{a}{c} x=a \text {. }
$$

Therefore, $x=\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$.