2.29. A plane divides the medians of the faces $ABC, ACD$, and $ADB$ of the tetrahedron $ABCD$, emanating from vertex $A$, in the ratios $2:1, 1:2$, and $4:1$, counting from the vertex. Let $P, Q$, and $R$ be the points of intersection of this plane with the lines $AB, AC$, and $AD$. Find the ratios $AP:PB$, $AQ:QC$, and $AR:RD$.
2.29. Let's restate the following problem. On the sides $AB$ and $AC$ of triangle $ABC$, points $L$ and $K$ are taken such that $AL : LB = m$ and $AK : KC = n$. Let $N$ be the intersection point of line $KL$ and median $AM$. We need to find the ratio $AN : NM$. For this, consider points $S$ and $T$, where line $KL$ intersects line $BC$ and the line parallel to $BC$ through point $A$. Clearly, $AT : SB = AL : LB = m$ and $AT : SC = AK : KC = n$. Therefore, $AN : NM = AT : SM = 2AT : (SC + SB) = 2(SC : AT + SB : AT)^{-1} = 2mn / (m + n)$. Note that all the reasoning remains valid even when points $K$ and $L$ are taken on the extensions of the sides of the triangle; in this case, the numbers $m$ and $n$ are negative.

Now, suppose that $AP : PB = p$, $AQ : QC = q$, and $AR : RD = r$. According to the problem, $2pq / (p + q) = 2$, $2qr / (q + r) = 1/2$, and $2pr / (p + r) = 4$. Solving this system of equations, we get $p = -4/5$, $q = 4/9$, and $r = 4/7$. The negativity of the number $p$ means that the given plane intersects not the edge $AB$, but its extension.