10 Given that point $G$ is the center of a regular tetrahedron $O A B C$ with edge length $a$, a plane passing through point $G$ intersects the rays $O A$, $O B$, and $O C$ at points $M$, $N$, and $P$ respectively. Find the minimum value of $O M+O N+O P$.
10 Draw a plane through $G$ parallel to plane $OBC$, intersecting $OA$ at $M'$, $PM$ at $Q$, and $NM$ at $Q'$. Then $Q Q' \parallel PN$, $OM' = \frac{1}{4} OA = \frac{a}{4} \cdot \frac{OM'}{OM} = \frac{PQ}{PM}$. Similarly, draw $SS' \parallel PM$, $TT' \parallel MN$, with $S$ on $MN$, $T$, and $S'$ on $NP$, and $T'$ on $PM$. (As shown in Figure 21), then $\frac{a / 4}{ON} = \frac{MS}{MN}$, $\frac{a / 4}{OP} = \frac{NT}{PN}$.
Since
$$
\frac{MS}{MN} = \frac{T'G}{MN} = \frac{T'G}{T'T} \times \frac{T'T}{MN} = \frac{T'Q}{T'P} \times \frac{T'P}{PM} = \frac{T'Q}{PM}.
$$

Therefore,
$$
\begin{array}{c}
\frac{a}{4}\left(\frac{1}{OM} + \frac{1}{ON} + \frac{1}{OP}\right) = \frac{PQ}{PM} + \frac{NT}{PN} + \frac{MS}{MN} = \frac{PQ}{PM} + \frac{MT'}{PM} + \frac{T'Q}{PM} = 1. \\
(OM + ON + OP)\left(\frac{1}{OM} + \frac{1}{ON} + \frac{1}{OP}\right) \geqslant 9.
\end{array}
$$

Therefore,
$$
OM + ON + OP \geqslant \frac{9}{4} a.
$$

The minimum value $\frac{9}{4} a$ is achieved when the plane $MNP \perp OG$.