Let $$\vec{\alpha} = (\lambda - 2)\vec{a} + \vec{b}$$ and $$\vec{\beta} = (4\lambda - 2)\vec{a} + 3\vec{b}$$, be two given vectors where vectors $$\vec{a}$$ and $$\vec{b}$$ are non-collinear. The value of $$\lambda$$ for which vectors $$\vec{\alpha}$$ and $$\vec{\beta}$$ are collinear, is:

We have two vectors defined in terms of the non-collinear base vectors $$\vec a$$ and $$\vec b$$:

$$\vec\alpha = (\lambda-2)\vec a + \vec b$$

$$\vec\beta = (4\lambda-2)\vec a + 3\vec b$$

Because $$\vec a$$ and $$\vec b$$ are not collinear, they form an independent pair. Any vector written as a linear combination of them is uniquely determined by its two scalar coefficients. Two such vectors will be collinear if and only if their respective coefficients are in the same ratio. In other words, for some real number $$k$$ we must have

$$\vec\alpha = k\,\vec\beta.$$

Equating the coefficients of $$\vec a$$ and $$\vec b$$ separately, we obtain

$$\dfrac{\text{coefficient of }\vec a\text{ in }\vec\alpha}{\text{coefficient of }\vec a\text{ in }\vec\beta} \;=\; \dfrac{\text{coefficient of }\vec b\text{ in }\vec\alpha}{\text{coefficient of }\vec b\text{ in }\vec\beta}.$$

Explicitly substituting the given coefficients, this condition becomes

$$\frac{\lambda-2}{4\lambda-2} = \frac{1}{3}.$$

Now we solve step by step. First cross-multiply:

$$3(\lambda-2) = 1\,(4\lambda-2).$$

Expand both sides:

$$3\lambda - 6 = 4\lambda - 2.$$

Bring all terms to one side to isolate $$\lambda$$:

$$3\lambda - 6 - 4\lambda + 2 = 0.$$

Combine like terms:

$$(3\lambda - 4\lambda) + (-6 + 2) = 0,$$

$$-\,\lambda - 4 = 0.$$

Multiply by $$-1$$ to make the coefficient positive:

$$\lambda + 4 = 0.$$

Therefore,

$$\lambda = -4.$$

Hence, the correct answer is Option A.