For an equilateral prism where the angle of prism $$A$$ is equal to $$60^\circ$$, when the light undergoes minimum deviation, the angle of incidence ($$i$$) becomes equal to the angle of emergence ($$e$$) and both are equal to the angle of prism divided by 2, i.e., $$i = e = \frac{A}{2}$$.

In this question, it is given that the angle of incidence is $$\frac{3}{4} \times A$$. When the light undergoes minimum deviation, for an equilateral prism, $$\frac{A}{2} = \frac{3}{4} \times A$$. Solving for $$A$$:

$$ \frac{A}{2} = \frac{3A}{4} $$

$$ \frac{1A}{2} = \frac{3A}{4} $$

This equation is incorrect, based on the information about minimum deviation. But for solving the refraction through a prism at minimum deviation and relating it to the velocity of light in the prism:

We use Snell's Law and the formula for refractive index. The refractive index $$n$$ of the glass can be given by:

$$ n = \frac{\sin\left(\frac{\delta + A}{2}\right)}{\sin\left(\frac{A}{2}\right)} $$

At minimum deviation, $$i = e = \frac{A}{2}$$ and the deviation $$\delta$$ is minimum. For an equilateral prism where $$A = 60^\circ$$:

$$ n = \frac{\sin\left(\frac{\delta + 60^\circ}{2}\right)}{\sin\left(30^\circ\right)} $$

$$ \sin(30^\circ) = 0.5 $$

Since the angle of incidence is $$\frac{3}{4}A = \frac{3}{4} \times 60^\circ = 45^\circ$$, and given that this is minimum deviation, let's skip forward with using the fact that it achieves symmetry at minimum deviation, we simplify calculations by using the angle condition:

From Snell's law at air-glass interface:

$$ \sin(i) = n \sin(r) $$

$$ \sin(45^\circ) = n \sin(30^\circ) $$

$$ \frac{\sqrt{2}}{2} = n \times 0.5 $$

$$ n = \frac{\sqrt{2}}{1} = \sqrt{2} $$

The velocity of light in the glass prism, $$v$$, is related to the velocity in vacuum, $$c$$, by the refractive index:

$$ v = \frac{c}{n} $$

Given that $$n = \sqrt{2}$$:

$$ v = \frac{c}{\sqrt{2}} $$

Thus, the answer is Option D:

$$ \frac{c}{\sqrt{2}} $$

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