If $$\vec{a}$$ and $$\vec{b}$$ are non-collinear vectors, then the value of $$\alpha$$ for which the vectors $$\vec{u} = (\alpha - 2)\vec{a} + \vec{b}$$ and $$\vec{v} = (2 + 3\alpha)\vec{a} - 3\vec{b}$$ are collinear is :

We are given that $$\vec{a}$$ and $$\vec{b}$$ are non-collinear vectors, meaning they are linearly independent. The vectors are defined as $$\vec{u} = (\alpha - 2)\vec{a} + \vec{b}$$ and $$\vec{v} = (2 + 3\alpha)\vec{a} - 3\vec{b}$$. We need to find the value of $$\alpha$$ such that $$\vec{u}$$ and $$\vec{v}$$ are collinear.

Since $$\vec{u}$$ and $$\vec{v}$$ are collinear, one must be a scalar multiple of the other. Therefore, we can write $$\vec{u} = k \vec{v}$$ for some scalar $$k$$. Substituting the expressions, we get:

$$(\alpha - 2)\vec{a} + \vec{b} = k \left[ (2 + 3\alpha)\vec{a} - 3\vec{b} \right]$$

Expanding the right-hand side:

$$(\alpha - 2)\vec{a} + \vec{b} = k(2 + 3\alpha)\vec{a} - 3k\vec{b}$$

Because $$\vec{a}$$ and $$\vec{b}$$ are linearly independent, the coefficients of $$\vec{a}$$ and $$\vec{b}$$ on both sides must be equal. This gives us two equations:

1. For $$\vec{a}$$: $$\alpha - 2 = k(2 + 3\alpha)$$

2. For $$\vec{b}$$: $$1 = -3k$$

Solving the second equation for $$k$$:

$$1 = -3k \implies k = -\frac{1}{3}$$

Substitute $$k = -\frac{1}{3}$$ into the first equation:

$$\alpha - 2 = \left(-\frac{1}{3}\right)(2 + 3\alpha)$$

Simplify the right-hand side:

$$\alpha - 2 = -\frac{1}{3} \times (2 + 3\alpha) = -\frac{2}{3} - \alpha$$

Now, solve for $$\alpha$$. Add $$\alpha$$ to both sides:

$$\alpha + \alpha - 2 = -\frac{2}{3}$$

$$2\alpha - 2 = -\frac{2}{3}$$

Add 2 to both sides:

$$2\alpha = -\frac{2}{3} + 2$$

$$2\alpha = -\frac{2}{3} + \frac{6}{3} = \frac{4}{3}$$

Divide both sides by 2:

$$\alpha = \frac{4}{3} \times \frac{1}{2} = \frac{4}{6} = \frac{2}{3}$$

Thus, $$\alpha = \frac{2}{3}$$. Now, verifying with the options:

A. $$\frac{3}{2}$$

B. $$\frac{2}{3}$$

C. $$-\frac{3}{2}$$

D. $$-\frac{2}{3}$$

The value $$\alpha = \frac{2}{3}$$ corresponds to option B. To confirm, substitute $$\alpha = \frac{2}{3}$$ into $$\vec{u}$$ and $$\vec{v}$$:

$$\vec{u} = \left(\frac{2}{3} - 2\right)\vec{a} + \vec{b} = \left(-\frac{4}{3}\right)\vec{a} + \vec{b}$$

$$\vec{v} = \left(2 + 3 \times \frac{2}{3}\right)\vec{a} - 3\vec{b} = (2 + 2)\vec{a} - 3\vec{b} = 4\vec{a} - 3\vec{b}$$

Check if $$\vec{u} = k \vec{v}$$ with $$k = -\frac{1}{3}$$:

$$k \vec{v} = -\frac{1}{3} (4\vec{a} - 3\vec{b}) = -\frac{4}{3}\vec{a} + \vec{b} = \vec{u}$$

This holds true, confirming that $$\vec{u}$$ and $$\vec{v}$$ are collinear when $$\alpha = \frac{2}{3}$$.

Hence, the correct answer is Option B.