1. We are given $\vec\alpha=(\lambda-2)\vec a+\vec b$ and $\vec\beta=(4\lambda-2)\vec a+3\vec b.$

   Also, $\vec a$ and $\vec b$ are non-collinear, so they are linearly independent.

2. For $\vec\alpha$ and $\vec\beta$ to be collinear, one must be a scalar multiple of the other. So there exists some scalar $k$ such that $\vec\beta=k\vec\alpha.$

3. Substitute the expressions: $(4\lambda-2)\vec a+3\vec b = k\big((\lambda-2)\vec a+\vec b\big).$

4. Compare coefficients of $\vec a$ and $\vec b$.

   Since $\vec a$ and $\vec b$ are independent, $4\lambda-2 = k(\lambda-2),$ $3 = k.$

5. Put $k=3$ into the first equation: $4\lambda-2 = 3(\lambda-2).$

   Solve: $4\lambda-2 = 3\lambda-6$ $4\lambda-3\lambda = -6+2$ $\lambda = -4.$

6. Hence the required value is $\boxed{-4}.$

7. Checking options:

   • A: $4$ ❌
   • B: $3$ ❌
   • C: $-3$ ❌
   • D: $-4$ ✅

Therefore, the correct option is D.