If $$\vec{a}, \vec{b}, \vec{c}$$ are non-coplanar vectors and $$\lambda$$ is a real number, then the vectors $$\vec{a} + 2\vec{b} + 3\vec{c}, \lambda \vec{b} + 4\vec{c}$$ and $$(2\lambda - 1)\vec{c}$$ are non-coplanar for

Since $$\vec a,\vec b,\vec c$$ are non-coplanar,

$$[\vec a,\vec b,\vec c]\ne0$$

Let

$$\vec p=\vec a+2\vec b+3\vec c$$

$$\vec q=\lambda\vec b+4\vec c$$

$$\vec r=(2\lambda-1)\vec c$$

The vectors $$\vec p,\vec q,\vec r$$ will be non-coplanar if

$$[\vec p,\vec q,\vec r]\ne0$$

Now,

$$[\vec p,\vec q,\vec r]=[\vec a+2\vec b+3\vec c,\lambda\vec b+4\vec c,(2\lambda-1)\vec c]$$

Using linearity of scalar triple product,

$$=(2\lambda-1)[\vec a+2\vec b+3\vec c,\lambda\vec b+4\vec c,\vec c]$$

Again expanding,

$$=(2\lambda-1)\left(\lambda[\vec a,\vec b,\vec c]+4[\vec a,\vec c,\vec c]+2\lambda[\vec b,\vec b,\vec c]+8[\vec b,\vec c,\vec c]+3\lambda[\vec c,\vec b,\vec c]+12[\vec c,\vec c,\vec c]\right)$$

All terms containing two equal vectors vanish.

Therefore,

$$[\vec p,\vec q,\vec r]=(2\lambda-1)\lambda[\vec a,\vec b,\vec c]$$

Since

$$[\vec a,\vec b,\vec c]\ne0$$

we must have

$$(2\lambda-1)\lambda\ne0$$

Hence,

$$\lambda\ne0,\qquad \lambda\ne\frac12$$

Therefore, the vectors are non-coplanar for $$\boxed{\lambda\in\mathbb{R}\setminus\left\{0,\frac{1}{2}\right\}}$$