In the $∆OAB, M$is the mid-point of $AB, C$ is a point on $OM$, such that $2OC=CM.X$ is a point on the side $OB$ such that $OX=2XB$. The line $XC$ is produced to meet $OA$ in $Y$. Then, $\frac{OY}{YA}$ is equal to

$$\begin{gathered} OA=a, OB=b \\ \therefore OM=\frac{a+b}{2} \\ \therefore OC=\frac{a+b}{6} \\ OX=\frac{2}{3}b \end{gathered}$$

Let    $\frac{OY}{YA}=\lambda$

$\therefore \frac{OY}{YA}=\frac{\lambda }{\lambda +1}a$

Now points $Y, C$ and $X$ are collinear.

$$\begin{gathered} \therefore YC=mCX \\ \therefore \frac{a+b}{6}-\frac{\lambda }{\lambda +1}a=m\frac{2b}{3}-m\frac{a+b}{6} \end{gathered}$$

Comparing coefficients of $a$ and $b$

$\therefore \frac{1}{6}-\frac{\lambda }{\lambda +1}=-\frac{m}{6}$

and $\frac{1}{6}=\frac{2m}{3}-\frac{m}{6}$

$\therefore m=\frac{1}{3}$ and $\lambda =\frac{2}{7}$