A bookshelf contains 6 distinct books of Mathematics and 5 distinct...

Total number of ways to choose $6$ books from $11$ books is $^{11}C_{6} = 462$.

Let $M$ and $P$ be the number of Mathematics and Physics books chosen respectively. Since $6$ books are chosen, $M + P = 6$.

The random variable $X$ is given by $X = |M - P| = |2M - 6|$.

The possible selections of $(M, P)$, the corresponding values of $X$, and the number of ways are as follows:

For $M=1, P=5$: $X = |1 - 5| = 4$, Number of ways $= ^{6}C_{1} \times ^{5}C_{5} = 6 \times 1 = 6$

For $M=2, P=4$: $X = |2 - 4| = 2$, Number of ways $= ^{6}C_{2} \times ^{5}C_{4} = 15 \times 5 = 75$

For $M=3, P=3$: $X = |3 - 3| = 0$, Number of ways $= ^{6}C_{3} \times ^{5}C_{3} = 20 \times 10 = 200$

For $M=4, P=2$: $X = |4 - 2| = 2$, Number of ways $= ^{6}C_{4} \times ^{5}C_{2} = 15 \times 10 = 150$

For $M=5, P=1$: $X = |5 - 1| = 4$, Number of ways $= ^{6}C_{5} \times ^{5}C_{1} = 6 \times 5 = 30$

For $M=6, P=0$: $X = |6 - 0| = 6$, Number of ways $= ^{6}C_{6} \times ^{5}C_{0} = 1 \times 1 = 1$

The mean $\alpha$ of the random variable $X$ is given by:

$\alpha = \dfrac{\sum X_i f_i}{\sum f_i}$

$\alpha = \dfrac{4(6) + 2(75) + 0(200) + 2(150) + 4(30) + 6(1)}{462}$

$\alpha = \dfrac{24 + 150 + 0 + 300 + 120 + 6}{462}$

$\alpha = \dfrac{600}{462} = \dfrac{100}{77}$

Therefore, the value of $77\alpha$ is:

$77\alpha = 77 \times \dfrac{100}{77} = 100$

Answer: $100$