67. Find three numbers such that the largest exceeds the middle by a given part $\left(\frac{1}{3}\right)$ of the smallest, the middle exceeds the smallest by a given part $\left(\frac{1}{3}\right)$ of the largest, and the smallest exceeds the number 10 by a given part $\left(\frac{1}{3}\right)$ of the middle number.
67. The task is reduced to solving the system:

$$
\begin{gathered}
x-y=\frac{1}{3} z ; \quad y-z=\frac{1}{3} x ; \quad z-10=\frac{1}{3} y \\
\quad x=45 ; y=37 \frac{1}{2} ; z=22 \frac{1}{2}
\end{gathered}
$$

$$
\text { from which }
$$