[Solution] Draw the medians $A N, B M$ on $A C, B C$ intersecting at $G$, and draw $M J \perp R S$, and $G L / / R S, B K / / R S$.
Thus, 
$$
M J=\frac{1}{2}(A D+F C)
$$
$$
=\frac{1}{2}(10+24)=17,
$$

Therefore,
$$
\begin{aligned}
M K & =M J-K J \\
& =M J-B E \\
& =17-6=11 .
\end{aligned}
$$

Since $M G=\frac{1}{3} M B$, and $L G / / K B$, it follows that $M L=\frac{1}{3} M K$.
Thus,
$$
\begin{aligned}
G H & =L J=M J-M L=M J-\frac{1}{3} M K \\
& =17-\frac{1}{3} \cdot 11=\frac{40}{3} .
\end{aligned}
$$

Therefore, the answer is $(A)$.
Note The physical significance of this problem is evident.
In fact, $G H=\frac{1}{3}(A D+B E+C F)$.
$17 \cdot 66$ Line $RS$ does not intersect $\triangle ABC$, and perpendiculars are drawn from the vertices $A, B, C$ to $RS$, with the corresponding feet of the perpendiculars being $D, E, F$, respectively. Also, $AD=10, BE=6, CF=24$. If $H$ is the foot of the perpendicular from the centroid $G$ of $\triangle ABC$ to $RS$, and $x$ represents the length of segment $GH$, then $x$ is
(A) $\frac{40}{3}$.
(B) 16 .
(C) $\frac{56}{3}$.
(D) $\frac{80}{3}$.
(E) cannot be determined.
(18th American High School Mathematics Examination, 1967)