The sum of all values of $$\alpha$$, for which the points whose position vectors are $$\hat{i} - 2\hat{j} + 3\hat{k}$$, $$2\hat{i} - 3\hat{j} + 4\hat{k}$$, $$(\alpha+1)\hat{i} + 2\hat{k}$$ and $$9\hat{i} + (\alpha-8)\hat{j} + 6\hat{k}$$ are coplanar, is equal to

We begin by letting the position vectors of the four points be A(1,-2,3), B(2,-3,4), C(α+1,0,2), and D(9,α-8,6).

Since coplanarity of A, B, C, D requires that the scalar triple product of AB, AC, and AD is zero, we compute:

$$\vec{AB} = (1,-1,1)$$, $$\vec{AC} = (\alpha, 2, -1)$$, $$\vec{AD} = (8, \alpha-6, 3)$$.

Next, we set up the determinant corresponding to the scalar triple product:
$$ \begin{vmatrix} 1 & -1 & 1 \\ \alpha & 2 & -1 \\ 8 & \alpha-6 & 3 \end{vmatrix} = 0 $$

Expanding this determinant along the first row gives:

$$= 1(6+\alpha-6) - (-1)(3\alpha+8) + 1(\alpha^2-6\alpha-16)$$

Combining like terms, we obtain:

$$= \alpha + 3\alpha + 8 + \alpha^2 - 6\alpha - 16$$

Simplifying further leads to

$$= \alpha^2 - 2\alpha - 8 = 0$$

Factoring this quadratic expression yields

$$(\alpha - 4)(\alpha + 2) = 0$$

Therefore, $$\alpha = 4$$ or $$\alpha = -2$$, and summing these values gives $$4 + (-2) = 2$$.

Hence, the correct answer is 2.