The number of all possible positive integral values of $$\alpha$$ for which the roots of the quadratic equation $$6x^2 - 11x + \alpha = 0$$ are rational numbers is:

We are given the quadratic equation $$6x^2 - 11x + \alpha = 0$$ and need to find all positive integral values of $$\alpha$$ for which both roots are rational.

For a quadratic $$ax^2 + bx + c = 0$$, the roots are rational if and only if the discriminant $$D = b^2 - 4ac$$ is a perfect square (and non-negative).

Here $$a = 6$$, $$b = -11$$, $$c = \alpha$$, so:

$$D = (-11)^2 - 4(6)(\alpha) = 121 - 24\alpha$$

For the roots to be real, we need $$D \geq 0$$:

$$121 - 24\alpha \geq 0$$

$$\alpha \leq \frac{121}{24} \approx 5.04$$

Since $$\alpha$$ must be a positive integer, the possible values are $$\alpha \in \{1, 2, 3, 4, 5\}$$.

Now we check which values make $$D$$ a perfect square:

For $$\alpha = 1$$: $$D = 121 - 24 = 97$$ (not a perfect square) ✘

For $$\alpha = 2$$: $$D = 121 - 48 = 73$$ (not a perfect square) ✘

For $$\alpha = 3$$: $$D = 121 - 72 = 49 = 7^2$$ ✔

For $$\alpha = 4$$: $$D = 121 - 96 = 25 = 5^2$$ ✔

For $$\alpha = 5$$: $$D = 121 - 120 = 1 = 1^2$$ ✔

Let us verify the roots for each valid value:

For $$\alpha = 3$$: roots are $$x = \frac{11 \pm 7}{12}$$, giving $$x = \frac{3}{2}$$ and $$x = \frac{1}{3}$$ (both rational) ✔

For $$\alpha = 4$$: roots are $$x = \frac{11 \pm 5}{12}$$, giving $$x = \frac{4}{3}$$ and $$x = \frac{1}{2}$$ (both rational) ✔

For $$\alpha = 5$$: roots are $$x = \frac{11 \pm 1}{12}$$, giving $$x = 1$$ and $$x = \frac{5}{6}$$ (both rational) ✔

Therefore, there are $$3$$ positive integral values of $$\alpha$$ (namely $$3, 4, 5$$) for which the roots are rational.

The correct answer is Option B: 3.