148. A plane intersecting the surface of a triangular pyramid divides the medians of the faces emanating from one vertex in the ratios $2: 1, 1: 2, 4: 1$ respectively (counting from the vertex). In what ratio does this plane divide the volume of the pyramid?
148. Let's first solve the following problem. In $\triangle ABC$, points $L$ and $K$ are taken on sides $AB$ and $AC$ such that $\frac{|AL|}{|LB|}=m, \quad \frac{|AK|}{|KC|}=n$. In what ratio does the line $KL$ divide the median $AM$?

Let $N$ be the intersection point of $KL$ and $AM$, $Q$ be the intersection point of $KL$ and $BC$, and $P$ be the intersection point of $KL$ and the line parallel to $BC$ passing through $A$. Suppose $|BC|=2a, |QC|=b, |AP|=c, n>m$. Then, from the similarity of the corresponding triangles, we have: $\frac{c}{b}=n, \frac{c}{b+2a}=m$, from which $\frac{|AN|}{|NM|}=\frac{c}{b+a}=\frac{2mn}{m+n}$.

Now let $m, n$, and $p$ be the ratios in which the plane divides the edges $AB, AC$, and $AD$. To determine these, we have the system

$$
\frac{2mn}{m+n}=2, \quad \frac{2np}{n+p}=\frac{1}{2}, \quad \frac{2pm}{p+m}=4
$$

from which

$$
m=-\frac{4}{5}, \quad n=\frac{4}{9}, \quad p=\frac{4}{7}
$$

The fact that $-1<m<0$ means that point $L$ lies on the extension of $AB$ beyond point $A$, i.e., our plane intersects the edges $AC, AD, BC$, and $BD$. Further, by determining the ratios in which the edges $BC$ and $BD$ are divided (we get $\frac{5}{7}$ and $\frac{5}{9}$), we find the answer: $\frac{7123}{16901}$.