Solve this following
Question:

Let $\lambda \neq 0$ be in $\mathrm{R}$. If $\alpha$ and $\beta$ are the roots of the equation, $x^{2}-x+2 \lambda=0$ and $\alpha$ and $\gamma$ are the roots of the equation, $3 x^{2}-10 x+27 \lambda=0$,

then $\frac{\beta \gamma}{\lambda}$ is equal to :

1. 36

2. 27

3. 9

4. 18

Correct Option: , 4

Solution:

$\alpha+\beta=1, \quad \alpha \beta=2 \lambda$

$\alpha+\beta=\frac{10}{3}, \quad \alpha \gamma=\frac{27 \lambda}{3}=9 \lambda$

$\gamma-\beta=\frac{7}{3}$

$\frac{\gamma}{\beta}=\frac{9}{2} \Rightarrow \gamma=\frac{9}{2} \beta=\frac{9}{2} \times \frac{2}{3} \Rightarrow \gamma=3$

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

$\frac{9}{2} \beta=\frac{7}{3} \Rightarrow \beta=\frac{2}{3}$

$\alpha=1-\frac{2}{3}=\frac{1}{3}$

$2 \lambda=\frac{2}{9} \Rightarrow \lambda=\frac{1}{9}$

$\frac{\beta \gamma}{\lambda}=\frac{\frac{2}{3} \times 3}{\frac{1}{9}}=18$