If $$\vec{a}$$ and $$\vec{b}$$ are two vectors such that $$|\vec{a}| = 2$$ and $$|\vec{b}| = 3$$, then the maximum value of $$3|3\vec{a} + 2\vec{b}| + 4|3\vec{a} - 2\vec{b}|$$ is :

Let the angle between $$\vec{a}$$ and $$\vec{b}$$ be $$\theta$$. We have $$|\vec{a}| = 2$$, $$|\vec{b}| = 3$$, so $$\vec{a} \cdot \vec{b} = 6\cos\theta$$.

Computing the magnitudes:

$$|3\vec{a} + 2\vec{b}|^2 = 9(4) + 12(6\cos\theta) + 4(9) = 72 + 72\cos\theta = 144\cos^2\frac{\theta}{2}$$

So $$|3\vec{a} + 2\vec{b}| = 12\left|\cos\frac{\theta}{2}\right|$$.

$$|3\vec{a} - 2\vec{b}|^2 = 9(4) - 12(6\cos\theta) + 4(9) = 72 - 72\cos\theta = 144\sin^2\frac{\theta}{2}$$

So $$|3\vec{a} - 2\vec{b}| = 12\left|\sin\frac{\theta}{2}\right|$$.

The expression becomes (taking $$\theta \in [0, \pi]$$ so that $$\cos(\theta/2) \geq 0$$ and $$\sin(\theta/2) \geq 0$$):

$$3|3\vec{a} + 2\vec{b}| + 4|3\vec{a} - 2\vec{b}| = 36\cos\frac{\theta}{2} + 48\sin\frac{\theta}{2}$$

This is of the form $$A\cos\phi + B\sin\phi$$, whose maximum value is $$\sqrt{A^2 + B^2}$$:

$$\sqrt{36^2 + 48^2} = \sqrt{1296 + 2304} = \sqrt{3600} = 60$$

Hence, the correct answer is Option 3.