$$\text{If }7 = 5 + \frac{1}{7}(5+\alpha) + \frac{1}{7^2}(5+2\alpha)+ \frac{1}{7^3}(5+3\alpha) + \cdots + \infty,\text{ then the value of } \alpha \text{ is:}$$

$$7=5+\dfrac{1}{7}(5+\alpha)+\dfrac{1}{7^2}(5+2\alpha)+\dfrac{1}{7^3}(5+3\alpha)+\cdots+\infty$$

$$7=5\left(1+\dfrac{1}{7}+\dfrac{1}{7^2}+\cdots+\infty\right)+\alpha\ \left(\dfrac{1}{7}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\cdots+\infty\ \right)$$

$$1+\dfrac{1}{7}+\dfrac{1}{7^2}+\cdots+\infty$$ is an infinite G.P with the common ratio of $$\dfrac{1}{7}$$.

Sum of the above G.P = $$\dfrac{1}{1-\dfrac{1}{7}}=\dfrac{7}{6}$$

$$7=\left(5\times\dfrac{7}{6}\right)+\alpha\ \left(\dfrac{1}{7}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\cdots+\infty\ \right)$$

$$7-\dfrac{35}{6}=\alpha\ \left(\dfrac{1}{7}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\cdots+\infty\ \right)$$

$$\dfrac{7}{6}=\alpha\ \left(\dfrac{1}{7}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\cdots+\infty\ \right)$$

Now, $$S=\left(\dfrac{1}{7}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\cdots+\infty\right)$$ in the form of AGP where $$\left(1,\ 2,\ 3,\ \cdots,\ \infty\right)$$ in A.P and $$\left(\dfrac{1}{7},\ \dfrac{1}{7^2},\ \dfrac{1}{7^3},\ \cdots,\ \infty\right)$$ are in G.P. 

$$S=\dfrac{1}{7}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\cdots+\infty\ $$                                        $$\longrightarrow\ i$$

Multiply both sides by $$\dfrac{1}{7}$$. 

$$\dfrac{S}{7}=\dfrac{1}{7^2}+\dfrac{2}{7^3}+\dfrac{3}{7^4}+\cdots+\infty\ $$                       $$\longrightarrow\ ii$$

Subtract equation $$ii$$ from equation $$i$$, 

$$\dfrac{6S}{7}=\dfrac{1}{7}+\left(\dfrac{2}{7^2}-\dfrac{1}{7^2}\right)+\left(\dfrac{3}{7^3}-\dfrac{2}{7^3}\right)+\left(\dfrac{4}{7^4}-\dfrac{3}{7^4}\right)+\cdots+\infty$$

$$\dfrac{6S}{7}=\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+\dfrac{1}{7^4}+\cdots+\infty$$

$$\dfrac{6S}{7}=\dfrac{\dfrac{1}{7}}{\left(1-\dfrac{1}{7}\right)}$$

$$\dfrac{6S}{7}=\dfrac{1}{6}$$

$$S=\dfrac{7}{36}$$

Hence, $$\dfrac{7}{6}=\alpha\ \times\dfrac{7}{36}$$

$$\alpha\ =6$$

Hence, the value of $$\alpha$$ is 6. 
$$\therefore\ $$ The required answer is B.