Solution:
Let the plane meet the edges $DA$, $DB$ and $DC$ at points $P$, $Q$ and $R$, respectively. Set
$$\frac{DP}{DA} = x, \quad \frac{DQ}{DB} = y, \text{ and } \frac{DR}{DC} = z$$
Let $M$ be the midpoint of $AB$ and $L = DM \cap PQ$. It follows from the condition of the

problem that $\frac{DL}{DM} = \frac{1}{3}$. Therefore
$$\frac{S_{DLP}}{S_{DAM}} = \frac{DP \cdot DL}{DA \cdot DM} = \frac{x}{3} ; \quad \frac{S_{DLQ}}{S_{DMB}} = \frac{DL \cdot DQ}{DM \cdot DB} = \frac{y}{3}$$
Since $S_{DAM} = S_{DMB} = \frac{1}{2} S_{DAB}$ we conclude that
$$2xy = 2 \frac{DP \cdot DQ}{DA \cdot DB} = \frac{S_{DPQ}}{\frac{1}{2} S_{DAB}} = \frac{S_{DPL}}{S_{DAM}} + \frac{S_{DLQ}}{S_{DMB}} = \frac{x + y}{3}$$
i.e. $\frac{1}{x} + \frac{1}{y} = 6$. Analogously $\frac{1}{y} + \frac{1}{z} = 8$ and $\frac{1}{z} + \frac{1}{x} = 10$. Solving this system we obtain $x = \frac{1}{4}$, $y = \frac{1}{2}$ and $z = \frac{1}{6}$. Thus
$$\frac{V_{DPQR}}{V_{DABC}} = \frac{DP \cdot DQ \cdot DR}{DA \cdot DB \cdot DC} = xyz = \frac{1}{48}$$
and therefore the desired ratio equals $1:47$.