1. Interpret the given equations

We are given:

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

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

Also, $\lambda \ne 0$.

2. Use sum of roots

For the first equation, $x^2 - x + 2\lambda = 0$ Sum of roots: $\alpha + \beta = 1$

For the second equation, $3x^2 - 10x + 27\lambda = 0$ Sum of roots: $\alpha + \gamma = \frac{10}{3}$

Subtracting, $\gamma - \beta = \frac{10}{3} - 1 = \frac{7}{3}$

But this alone is not enough. We now use product of roots.

3. Use product of roots

For the first equation, $\alpha\beta = 2\lambda$

For the second equation, $\alpha\gamma = \frac{27\lambda}{3} = 9\lambda$

Now divide the second by the first: $\frac{\alpha\gamma}{\alpha\beta} = \frac{9\lambda}{2\lambda}$

Since $\lambda \ne 0$ and $\alpha \ne 0$ (otherwise $\alpha\beta=2\lambda\ne 0$ impossible), we get $\frac{\gamma}{\beta} = \frac{9}{2}$

So, $\gamma = \frac{9}{2}\beta$

4. Use the relation from sums

We have $\alpha = 1-\beta$ and also $\alpha = \frac{10}{3} - \gamma$

Equating: $1-\beta = \frac{10}{3} - \gamma$ $\gamma - \beta = \frac{7}{3}$

Substitute $\gamma = \frac{9}{2}\beta$: $\frac{9}{2}\beta - \beta = \frac{7}{3}$ $\frac{7}{2}\beta = \frac{7}{3}$ $\beta = \frac{2}{3}$

Hence, $\gamma = \frac{9}{2}\cdot \frac{2}{3} = 3$

5. Find $\lambda$

From $\alpha + \beta = 1$ we get $\alpha = 1 - \frac{2}{3} = \frac{1}{3}$

Now use $\alpha\beta = 2\lambda$ $\frac{1}{3}\cdot \frac{2}{3} = 2\lambda$ $\frac{2}{9} = 2\lambda$ $\lambda = \frac{1}{9}$

6. Compute the required value

$\frac{\beta\gamma}{\lambda} = \frac{\left(\frac{2}{3}\right)(3)}{1/9} = \frac{2}{1/9} = 18$

7. Check options

The value is $18$ So the correct option is D.