Let λ ≠ 0 be a real number. Let α, β be the roots of the equation...

Given $\alpha$ and $\beta$ are roots of the below equation

$14{x}^{2} - 31x + 3\lambda = 0$

So, $\alpha + \beta = \dfrac{31}{14} \ldots \left(1\right)$

And, $\alpha\beta = \dfrac{3\lambda}{14} \ldots \left(2\right)$

Given $\alpha$ and $\gamma$ are roots of the below equation

$35{x}^{2} - 53x + 4\lambda = 0$

So, $\alpha + \gamma = \dfrac{53}{35} \ldots \left(3\right)$

And, $\alpha\gamma = \dfrac{4\lambda}{35} \ldots \left(4\right)$

On dividing equation\left(2\right)$and equation \left(4\right)$,

$\dfrac{\beta}{\gamma} = \dfrac{3 \times 35}{4 \times 14} = \dfrac{15}{8} ⇒ \beta = \dfrac{15}{8}\gamma$

On subtracting equation \left(1\right)$and equation \left(3\right)$,

$⇒ \beta - \gamma = \dfrac{31}{14} - \dfrac{53}{35} = \dfrac{155 - 106}{70} = \dfrac{7}{10}$

$⇒ \dfrac{15}{8}\gamma - \gamma = \dfrac{7}{10} ⇒ \gamma = \dfrac{4}{5}$

$⇒ \beta = \dfrac{15}{8} \times \dfrac{4}{5} = \dfrac{3}{2}$

So, $\alpha = \dfrac{31}{14} - \beta = \dfrac{31}{14} - \dfrac{3}{2} = \dfrac{5}{7}$

And $\lambda = \dfrac{14}{3}\alpha\beta = \dfrac{14}{3} \times \dfrac{5}{7} \times \dfrac{3}{2} = 5$

So, the sum of the roots of the required equation,

$\dfrac{3\alpha}{\beta} + \dfrac{4\alpha}{\gamma} = \left(\dfrac{3\alpha\gamma + 4\alpha\beta}{\beta\gamma}\right)$

$= \dfrac{\left(3 \times \dfrac{4\lambda}{35} + 4 \times \dfrac{3\lambda}{14}\right)}{\beta\gamma} = \dfrac{12\lambda\left(14 + 35\right)}{14 \times 35\beta\gamma}$

$= \dfrac{49 \times 12 \times 5}{490 \times \dfrac{3}{2} \times \dfrac{4}{5}} = 5$

And the product of the roots of the required equation,
$= \dfrac{3\alpha}{\beta} \times \dfrac{4\alpha}{\gamma} = \dfrac{12{\alpha}^{2}}{\beta\gamma} = \dfrac{12 \times \dfrac{25}{49}}{\dfrac{3}{2} \times \dfrac{4}{5}} = \dfrac{250}{49}$

So, the required equation is

${x}^{2} - ( \text{sum of roots} ) x + \text{product of roots} = 0$

$⇒ {x}^{2} - 5x + \dfrac{250}{49} = 0$

$⇒ 49{x}^{2} - 245x + 250 = 0$