Let $$\vec{a}, \vec{b}$$ and $$\vec{c}$$ be three non-zero non-coplanar vectors. Let the position vectors of four points $$A$$, $$B$$, $$C$$ and $$D$$ be $$\vec{a} - \vec{b} + \vec{c}$$, $$\lambda\vec{a} - 3\vec{b} + 4\vec{c}$$, $$-\vec{a} + 2\vec{b} - 3\vec{c}$$ and $$2\vec{a} - 4\vec{b} + 6\vec{c}$$ respectively. If $$\vec{AB}$$, $$\vec{AC}$$ and $$\vec{AD}$$ are coplanar, then $$\lambda$$ is:

Let the position vectors of the points $$A$$, $$B$$, $$C$$, and $$D$$ be denoted as $$\vec{OA}$$, $$\vec{OB}$$, $$\vec{OC}$$, and $$\vec{OD}$$ respectively. We first find the vectors $$\vec{AB}$$, $$\vec{AC}$$, and $$\vec{AD}$$:

$$\vec{AB} = \vec{OB} - \vec{OA} = (\lambda - 1)\vec{a} - 2\vec{b} + 3\vec{c}$$

$$\vec{AC} = \vec{OC} - \vec{OA} = -2\vec{a} + 3\vec{b} - 4\vec{c}$$

$$\vec{AD} = \vec{OD} - \vec{OA} = \vec{a} - 3\vec{b} + 5\vec{c}$$

Since the vectors $$\vec{AB}$$, $$\vec{AC}$$, and $$\vec{AD}$$ are coplanar and the base vectors $$\vec{a}$$, $$\vec{b}$$, and $$\vec{c}$$ are non-coplanar, the scalar triple product $$[\vec{AB} \quad \vec{AC} \quad \vec{AD}]$$ must equal zero. This is equivalent to setting the determinant of their coefficients to zero:

$$\begin{vmatrix} \lambda - 1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{vmatrix} = 0$$

Expanding the determinant along the first row:

$$(\lambda - 1) \begin{vmatrix} 3 & -4 \\ -3 & 5 \end{vmatrix} - (-2) \begin{vmatrix} -2 & -4 \\ 1 & 5 \end{vmatrix} + 3 \begin{vmatrix} -2 & 3 \\ 1 & -3 \end{vmatrix} = 0$$

Evaluating each minor:

- $$\begin{vmatrix} 3 & -4 \\ -3 & 5 \end{vmatrix} = (3)(5) - (-4)(-3) = 15 - 12 = 3$$

- $$\begin{vmatrix} -2 & -4 \\ 1 & 5 \end{vmatrix} = (-2)(5) - (-4)(1) = -10 + 4 = -6$$

- $$\begin{vmatrix} -2 & 3 \\ 1 & -3 \end{vmatrix} = (-2)(-3) - (3)(1) = 6 - 3 = 3$$

Substituting these values back into the expanded equation:

$$3(\lambda - 1) + 2(-6) + 3(3) = 0$$

$$3\lambda - 3 - 12 + 9 = 0$$

$$3\lambda - 6 = 0$$

$$3\lambda = 6 \implies \lambda = 2$$

Conclusion:

The value of $$\lambda$$ is equal to 2.