If the four points, whose position vectors are $$3\hat{i} - 4\hat{j} + 2\hat{k}$$, $$\hat{i} + 2\hat{j} - \hat{k}$$, $$-2\hat{i} - \hat{j} + 3\hat{k}$$ and $$5\hat{i} - 2\alpha\hat{j} + 4\hat{k}$$ are coplanar, then $$\alpha$$ is equal to

Given four points with position vectors: $$A = 3\hat{i} - 4\hat{j} + 2\hat{k}$$, $$B = \hat{i} + 2\hat{j} - \hat{k}$$, $$C = -2\hat{i} - \hat{j} + 3\hat{k}$$, $$D = 5\hat{i} - 2\alpha\hat{j} + 4\hat{k}$$.

For coplanarity, $$[\vec{AB}, \vec{AC}, \vec{AD}] = 0$$.

Compute the vectors.

$$\vec{AB} = B - A = (-2, 6, -3)$$

$$\vec{AC} = C - A = (-5, 3, 1)$$

$$\vec{AD} = D - A = (2, -2\alpha + 4, 2)$$

Set the scalar triple product to zero.

$$\begin{vmatrix} -2 & 6 & -3 \\ -5 & 3 & 1 \\ 2 & 4-2\alpha & 2 \end{vmatrix} = 0$$

Expanding along the first row:

$$-2[3(2) - 1(4-2\alpha)] - 6[(-5)(2) - 1(2)] + (-3)[(-5)(4-2\alpha) - 3(2)] = 0$$

$$-2[6 - 4 + 2\alpha] - 6[-10 - 2] - 3[-20 + 10\alpha - 6] = 0$$

$$-2[2 + 2\alpha] - 6[-12] - 3[-26 + 10\alpha] = 0$$

$$-4 - 4\alpha + 72 + 78 - 30\alpha = 0$$

$$146 - 34\alpha = 0$$

$$\alpha = \frac{146}{34} = \frac{73}{17}$$

Therefore, the correct answer is Option A: $$\mathbf{\frac{73}{17}}$$.