1. Given equations and common root

We are given: $14x^2-31x+3\lambda=0$ with roots $\alpha,\beta$, and $35x^2-53x+4\lambda=0$ with roots $\alpha,\gamma$.

Since $\alpha$ is a common root, it satisfies both equations: $14\alpha^2-31\alpha+3\lambda=0 \quad ...(1)$ $35\alpha^2-53\alpha+4\lambda=0 \quad ...(2)$

2. Eliminate $\lambda$ to find $\alpha$

Multiply (1) by $4$ and (2) by $3$: $56\alpha^2-124\alpha+12\lambda=0$ $105\alpha^2-159\alpha+12\lambda=0$

Subtracting, $(105-56)\alpha^2-(159-124)\alpha=0$ $49\alpha^2-35\alpha=0$ $7\alpha(7\alpha-5)=0$

So, $\alpha=0 \quad \text{or} \quad \alpha=\frac57$

But if $\alpha=0$, then from (1), $3\lambda=0\Rightarrow \lambda=0$, which is not allowed since $\lambda\ne 0$.

Hence, $\boxed{\alpha=\frac57}$

3. Find $\beta$ using product of roots of first equation

For $14x^2-31x+3\lambda=0,$ product of roots is $\alpha\beta=\frac{3\lambda}{14}$

Substitute $\alpha=\frac57$: $\frac57\beta=\frac{3\lambda}{14}$ $\beta=\frac{3\lambda}{10}$

4. Find $\gamma$ using product of roots of second equation

For $35x^2-53x+4\lambda=0,$ product of roots is $\alpha\gamma=\frac{4\lambda}{35}$

Substitute $\alpha=\frac57$: $\frac57\gamma=\frac{4\lambda}{35}$ $\gamma=\frac{4\lambda}{25}$

5. Compute the required new roots

First root: $\frac{3\alpha}{\beta}=\frac{3\cdot \frac57}{\frac{3\lambda}{10}}=\frac{15}{7}\cdot \frac{10}{3\lambda}=\frac{50}{7\lambda}$

Second root: $\frac{4\alpha}{\gamma}=\frac{4\cdot \frac57}{\frac{4\lambda}{25}}=\frac{20}{7}\cdot \frac{25}{4\lambda}=\frac{125}{7\lambda}$

So the required quadratic has roots: $r_1=\frac{50}{7\lambda}, \qquad r_2=\frac{125}{7\lambda}$

6. Find $\lambda$ from sum of roots relation

For the first equation, $\alpha+\beta=\frac{31}{14}$

Substitute $\alpha=\frac57$: $\frac57+\beta=\frac{31}{14}$ $\beta=\frac{31}{14}-\frac{10}{14}=\frac{21}{14}=\frac32$

But we also found: $\beta=\frac{3\lambda}{10}$

Therefore, $\frac{3\lambda}{10}=\frac32$ $\lambda=5$

Thus the new roots are: $r_1=\frac{50}{7\cdot 5}=\frac{10}{7}, \qquad r_2=\frac{125}{7\cdot 5}=\frac{25}{7}$

7. Form the quadratic equation

Sum of roots: $r_1+r_2=\frac{10}{7}+\frac{25}{7}=\frac{35}{7}=5$

Product of roots: $r_1r_2=\frac{10}{7}\cdot \frac{25}{7}=\frac{250}{49}$

Hence the equation is: $x^2-5x+\frac{250}{49}=0$

Multiplying by $49$: $\boxed{49x^2-245x+250=0}$

8. Check options

This matches:

Option B: $49x^2-245x+250=0$