If the line $$\alpha x + 2y = 1$$, where $$\alpha \in R $$, does not meet the hyperbola $$x^{2}-9y^{2}=9$$, then a possible value of $$\alpha$$ is:

The hyperbola is $$x^{2}-9y^{2}=9$$.

The given line is $$\alpha x + 2y = 1$$, where $$\alpha \in \mathbb{R}$$.
Rewrite the line in slope-intercept form so that $$y$$ can be substituted:

$$2y = 1 - \alpha x \; \Rightarrow \; y = \frac{1-\alpha x}{2}$$

Substitute this expression for $$y$$ into the hyperbola:

$$x^{2}-9\left(\frac{1-\alpha x}{2}\right)^{2}=9$$

Multiply every term by 4 to clear the denominator:

$$4x^{2}-9\left(1-\alpha x\right)^{2}=36$$

Bring every term to the left-hand side:

$$4x^{2}-9\left(1-\alpha x\right)^{2}-36=0$$

Expand the square $$\left(1-\alpha x\right)^{2}=1-2\alpha x+\alpha^{2}x^{2}$$ and simplify:

$$4x^{2}-9\bigl(1-2\alpha x+\alpha^{2}x^{2}\bigr)-36=0$$
$$\Longrightarrow \; 4x^{2}-9+18\alpha x-9\alpha^{2}x^{2}-36=0$$
$$\Longrightarrow \; (4-9\alpha^{2})x^{2}+18\alpha x-45=0$$

This is a quadratic in $$x$$ of the form $$Ax^{2}+Bx+C=0$$ with

$$A=4-9\alpha^{2}, \quad B=18\alpha, \quad C=-45$$

The line will not meet the hyperbola when this quadratic has no real roots, i.e. when its discriminant is negative:

$$\Delta = B^{2}-4AC \lt 0$$

Compute the discriminant:

$$\Delta = (18\alpha)^{2}-4(4-9\alpha^{2})(-45)$$
$$\;\;=324\alpha^{2}-4\bigl(-180+405\alpha^{2}\bigr)$$
$$\;\;=324\alpha^{2}+720-1620\alpha^{2}$$
$$\;\;=720-1296\alpha^{2}$$

Set $$\Delta \lt 0$$ to obtain the required inequality:

$$720-1296\alpha^{2} \lt 0$$
$$\Longrightarrow \; -1296\alpha^{2} \lt -720$$
$$\Longrightarrow \; \alpha^{2} \gt \frac{720}{1296}$$
$$\Longrightarrow \; \alpha^{2} \gt \frac{5}{9}$$

Taking square roots:

$$|\alpha| \gt \frac{\sqrt{5}}{3} \approx 0.745$$

Among the given options (0.5, 0.6, 0.7, 0.8), the only value satisfying $$|\alpha| \gt 0.745$$ is $$\alpha = 0.8$$.

Hence, a possible value of $$\alpha$$ is $$0.8$$  →  Option D.