10,11 |  |

Point $M$ lies on the edge $CD$ of the parallelepiped $ABCD A1 B1 C1 D1$, with $CM: MD=1: 2$. Construct the section of the parallelepiped by a plane passing through point $M$ and parallel to the lines $DB$ and $AC1$. In what ratio does this plane divide the diagonal $A1C$ of the parallelepiped?
Through the line $B D$, parallel to the cutting plane, the plane $A B C D$ is drawn, which has a common point $M$ with the cutting plane. Therefore, the cutting plane intersects the plane $A B C D$ along a line passing through point $M$ parallel to $B D$. Let $K$ be the point of intersection of this line with the diagonal $A C$ of the parallelogram $A B C D$, and $L$ be the center of the parallelogram. Then

$$
\frac{C K}{C L}=\frac{C M}{C D}=\frac{1}{3}, \frac{C K}{A C}=\frac{1}{6}
$$

Through the line $A C_1$, parallel to the cutting plane, the plane $A A_1 C_1 C$ is drawn, which has a common point $K$ with the cutting plane. Therefore, the cutting plane intersects the plane $A A_1 C_1 C$ along a line passing through point $K$ parallel to $A C_1$. Let $P$ be the point of intersection of this line with the diagonal $A_1 C$ of the parallelepiped $A B C D A_1 B_1 C_1 D_1$, and $O$ be the center of the parallelepiped. Then

$$
\frac{C P}{C O}=\frac{C K}{A C}=\frac{1}{6}, \frac{C P}{C A_1}=\frac{1}{12}
$$

Therefore, $\frac{C P}{P A_1}=\frac{1}{11}$.

## Answer

$1: 11$.