Let $$P_1 : \vec{r} \cdot (2\hat{i} + \hat{j} - 3\hat{k}) = 4$$ be a plane. Let $$P_2$$ be another plane which passes through the points $$(2, -3, 2)$$, $$(2, -2, -3)$$ and $$(1, -4, 2)$$. If the direction ratios of the line of intersection of $$P_1$$ and $$P_2$$ be $$16, \alpha, \beta$$, then the value of $$\alpha + \beta$$ is equal to ______.

Given $$P_1: 2x + y - 3z = 4$$ and $$P_2$$ passes through $$(2,-3,2)$$, $$(2,-2,-3)$$ and $$(1,-4,2)$$. Two vectors in the plane are $$\vec{v_1} = (2,-2,-3) - (2,-3,2) = (0, 1, -5)$$ and $$\vec{v_2} = (1,-4,2) - (2,-3,2) = (-1, -1, 0)$$.

Normal to $$P_2$$: $$\vec{n_2} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & -5 \\ -1 & -1 & 0 \end{vmatrix}$$ = $$\hat{i}(0 - 5) - \hat{j}(0 - 5) + \hat{k}(0 + 1)$$ = $$-5\hat{i} + 5\hat{j} + \hat{k}$$.

Hence $$P_2: -5(x-2) + 5(y+3) + (z-2) = 0$$, which simplifies to $$-5x + 10 + 5y + 15 + z - 2 = 0$$, or $$-5x + 5y + z + 23 = 0$$, i.e. $$5x - 5y - z = 23$$.

The direction of the line of intersection is given by $$\vec{n_1} \times \vec{n_2}$$ where $$\vec{n_1} = (2, 1, -3)$$ and $$\vec{n_2} = (5, -5, -1)$$: $$\vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 5 & -5 & -1 \end{vmatrix} = \hat{i}(-1 - 15) - \hat{j}(-2 + 15) + \hat{k}(-10 - 5) = -16\hat{i} - 13\hat{j} - 15\hat{k}$$, giving direction ratios $$(-16, -13, -15)$$ or equivalently $$(16, 13, 15)$$.

Given direction ratios are $$16, \alpha, \beta$$, so $$\alpha = 13$$ and $$\beta = 15$$. Hence $$\alpha + \beta = 13 + 15 = 28$$.

The answer is $$\boxed{28}$$.