If $$|\vec{a}| = 2$$, $$|\vec{b}| = 3$$ and $$|2\vec{a} - \vec{b}| = 5$$, then $$|2\vec{a} + \vec{b}|$$ equals:

We are given the magnitudes of vectors $$\vec{a}$$ and $$\vec{b}$$: $$|\vec{a}| = 2$$ and $$|\vec{b}| = 3$$. Also, the magnitude of $$2\vec{a} - \vec{b}$$ is given as $$|2\vec{a} - \vec{b}| = 5$$. We need to find the magnitude of $$|2\vec{a} + \vec{b}|$$.

Recall that for any vector $$\vec{v}$$, the square of its magnitude is equal to the dot product of the vector with itself: $$|\vec{v}|^2 = \vec{v} \cdot \vec{v}$$. We can use this property to solve the problem.

First, consider the given $$|2\vec{a} - \vec{b}| = 5$$. Squaring both sides gives: $$|2\vec{a} - \vec{b}|^2 = 5^2 = 25.$$

Now, expand the left side using the dot product: $$(2\vec{a} - \vec{b}) \cdot (2\vec{a} - \vec{b}) = (2\vec{a}) \cdot (2\vec{a}) - (2\vec{a}) \cdot \vec{b} - \vec{b} \cdot (2\vec{a}) + (-\vec{b}) \cdot (-\vec{b}).$$

Simplify this expression. Since the dot product is commutative and distributive, we have: $$(2\vec{a}) \cdot (2\vec{a}) = 4 (\vec{a} \cdot \vec{a}) = 4 |\vec{a}|^2,$$ $$(-\vec{b}) \cdot (-\vec{b}) = \vec{b} \cdot \vec{b} = |\vec{b}|^2,$$ and the cross terms: $$-(2\vec{a}) \cdot \vec{b} - \vec{b} \cdot (2\vec{a}) = -2(\vec{a} \cdot \vec{b}) - 2(\vec{b} \cdot \vec{a}).$$

Since the dot product is commutative, $$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$$, so the cross terms become: $$-2(\vec{a} \cdot \vec{b}) - 2(\vec{a} \cdot \vec{b}) = -4(\vec{a} \cdot \vec{b}).$$

Therefore, the expansion is: $$|2\vec{a} - \vec{b}|^2 = 4 |\vec{a}|^2 - 4 (\vec{a} \cdot \vec{b}) + |\vec{b}|^2.$$

Substitute the known magnitudes $$|\vec{a}| = 2$$ and $$|\vec{b}| = 3$$, so $$|\vec{a}|^2 = 4$$ and $$|\vec{b}|^2 = 9$$: $$4 \times 4 - 4 (\vec{a} \cdot \vec{b}) + 9 = 25.$$

Simplify the left side: $$16 - 4 (\vec{a} \cdot \vec{b}) + 9 = 25,$$ $$25 - 4 (\vec{a} \cdot \vec{b}) = 25.$$

Subtract 25 from both sides: $$-4 (\vec{a} \cdot \vec{b}) = 0,$$ so $$\vec{a} \cdot \vec{b} = 0.$$

This means the dot product of $$\vec{a}$$ and $$\vec{b}$$ is zero, indicating that the vectors are perpendicular.

Now, we need to find $$|2\vec{a} + \vec{b}|$$. Square this expression: $$|2\vec{a} + \vec{b}|^2 = (2\vec{a} + \vec{b}) \cdot (2\vec{a} + \vec{b}).$$

Expand the dot product: $$(2\vec{a}) \cdot (2\vec{a}) + (2\vec{a}) \cdot \vec{b} + \vec{b} \cdot (2\vec{a}) + \vec{b} \cdot \vec{b}.$$

Simplify each term: $$(2\vec{a}) \cdot (2\vec{a}) = 4 (\vec{a} \cdot \vec{a}) = 4 |\vec{a}|^2,$$ $$\vec{b} \cdot \vec{b} = |\vec{b}|^2,$$ and the cross terms: $$(2\vec{a}) \cdot \vec{b} + \vec{b} \cdot (2\vec{a}) = 2(\vec{a} \cdot \vec{b}) + 2(\vec{a} \cdot \vec{b}) = 4(\vec{a} \cdot \vec{b}),$$ since the dot product is commutative.

So, $$|2\vec{a} + \vec{b}|^2 = 4 |\vec{a}|^2 + 4 (\vec{a} \cdot \vec{b}) + |\vec{b}|^2.$$

Substitute the values $$|\vec{a}|^2 = 4$$, $$|\vec{b}|^2 = 9$$, and $$\vec{a} \cdot \vec{b} = 0$$: $$4 \times 4 + 4 \times 0 + 9 = 16 + 0 + 9 = 25.$$

Therefore, $$|2\vec{a} + \vec{b}| = \sqrt{25} = 5.$$

Hence, the correct answer is Option A.