We look at the first quadratic equation $$x^{2}-x+2\lambda=0$$ whose two roots are given to be $$\alpha$$ and $$\beta$$.
For a general quadratic $$ax^{2}+bx+c=0$$, the relation between the coefficients and the roots is
$$\text{Sum of roots}= -\dfrac{b}{a},\qquad \text{Product of roots}= \dfrac{c}{a}.$$
Here $$a=1,\; b=-1,\; c=2\lambda$$, so we obtain
$$\alpha+\beta=-\dfrac{-1}{1}=1,\qquad \alpha\beta=\dfrac{2\lambda}{1}=2\lambda.$$
Now consider the second quadratic equation $$3x^{2}-10x+27\lambda=0$$ whose roots are $$\alpha$$ and $$\gamma$$.
Using the same formulas with $$a=3,\; b=-10,\; c=27\lambda$$, we have
$$\alpha+\gamma=-\dfrac{-10}{3}=\dfrac{10}{3},\qquad \alpha\gamma=\dfrac{27\lambda}{3}=9\lambda.$$
Our goal is to find the value of $$\dfrac{\beta\gamma}{\lambda}$$.
First we eliminate $$\lambda$$ by relating the two products. We already have
$$\alpha\beta=2\lambda,\qquad \alpha\gamma=9\lambda.$$
Dividing the second product by the first gives
$$\dfrac{\alpha\gamma}{\alpha\beta}=\dfrac{9\lambda}{2\lambda}\; \Longrightarrow\; \gamma=\dfrac{9}{2}\,\beta.$$
(Here the common factor $$\alpha$$ cancels out because $$\alpha\neq0$$.)
Next we express $$\beta$$ and $$\gamma$$ in terms of $$\alpha$$ by using the two sum relations.
From $$\alpha+\beta=1$$ we get $$\beta=1-\alpha.$$
From $$\alpha+\gamma=\dfrac{10}{3}$$ we get $$\gamma=\dfrac{10}{3}-\alpha.$$
But we have also found above that $$\gamma=\dfrac{9}{2}\beta=\dfrac{9}{2}(1-\alpha).$$
Equating these two expressions for $$\gamma$$, we write
$$\dfrac{9}{2}(1-\alpha)=\dfrac{10}{3}-\alpha.$$
To clear denominators, multiply every term by 6:
$$6\left(\dfrac{9}{2}\right)(1-\alpha)=6\left(\dfrac{10}{3}-\alpha\right)$$
$$\; \Longrightarrow\; 27(1-\alpha)=20-6\alpha.$$
Expanding the left side gives
$$27-27\alpha=20-6\alpha.$$
Bringing all terms to one side:
$$27-27\alpha-20+6\alpha=0$$
$$\Longrightarrow 7-21\alpha=0$$
$$\Longrightarrow \alpha=\dfrac{7}{21}=\dfrac{1}{3}.$$
Now we immediately find
$$\beta=1-\alpha=1-\dfrac{1}{3}=\dfrac{2}{3},$$
$$\gamma=\dfrac{10}{3}-\alpha=\dfrac{10}{3}-\dfrac{1}{3}=3.$$
To determine $$\lambda$$ itself, we substitute into $$\alpha\beta=2\lambda$$:
$$\dfrac{1}{3}\cdot\dfrac{2}{3}=2\lambda \; \Longrightarrow\; \dfrac{2}{9}=2\lambda \; \Longrightarrow\; \lambda=\dfrac{1}{9}.$$
Finally, we compute the required ratio:
$$\dfrac{\beta\gamma}{\lambda}= \dfrac{\left(\dfrac{2}{3}\right)(3)}{\dfrac{1}{9}} = \dfrac{2}{\dfrac{1}{9}} = 2 \times 9 = 18.$$
Hence, the correct answer is Option B.
91- 6303239042
