We are given:

1. $\alpha,\beta$ are roots of $x^2-x+2\lambda=0$

2. $\alpha,\gamma$ are roots of $3x^2-10x+27\lambda=0$

We need to find $\frac{\beta\gamma}{\lambda}.$

1. Use Vieta's formulas for the first equation

For $x^2-x+2\lambda=0,$ if roots are $\alpha,\beta$, then $\alpha+\beta=1, \qquad \alpha\beta=2\lambda.$

Hence, $\beta=1-\alpha.$

2. Use Vieta's formulas for the second equation

For $3x^2-10x+27\lambda=0,$ if roots are $\alpha,\gamma$, then $\alpha+\gamma=\frac{10}{3}, \qquad \alpha\gamma=\frac{27\lambda}{3}=9\lambda.$

Hence, $\gamma=\frac{10}{3}-\alpha.$

3. Since $\alpha$ is common, it satisfies both equations

So $\alpha$ must satisfy: $\alpha^2-\alpha+2\lambda=0 \quad \text{and} \quad 3\alpha^2-10\alpha+27\lambda=0.$

From the first equation, $2\lambda=-\alpha^2+\alpha$ so $\lambda=\frac{\alpha-\alpha^2}{2}.$

Substitute into the second equation: $3\alpha^2-10\alpha+27\left(\frac{\alpha-\alpha^2}{2}\right)=0.$

Multiply by $2$: $6\alpha^2-20\alpha+27\alpha-27\alpha^2=0.$

$-21\alpha^2+7\alpha=0.$

$7\alpha(-3\alpha+1)=0.$

So, $\alpha=0 \quad \text{or} \quad \alpha=\frac13.$

But the question states $\lambda\ne 0$.

If $\alpha=0$, then from $\lambda=\frac{\alpha-\alpha^2}{2},$ we get $\lambda=0$, not allowed.

Therefore, $\alpha=\frac13.$

4. Find $\beta,\gamma,$ and $\lambda$

From $\alpha+\beta=1,$ $\beta=1-\frac13=\frac23.$

From $\alpha+\gamma=\frac{10}{3},$ $\gamma=\frac{10}{3}-\frac13=3.$

From $\alpha\beta=2\lambda,$ $\frac13\cdot\frac23=2\lambda$ $\frac{2}{9}=2\lambda$ $\lambda=\frac19.$

5. Compute the required value

$\frac{\beta\gamma}{\lambda}=\frac{\left(\frac23\right)(3)}{1/9}.$

First, $\left(\frac23\right)(3)=2.$

So, $\frac{\beta\gamma}{\lambda}=\frac{2}{1/9}=18.$

Final Answer

$\boxed{18}$

The derived answer matches the stored correct answer.