A line $$l$$ passing through origin is perpendicular to the lines
$$l_1: \vec{r} = (3 + t)\hat{i} + (-1 + 2t)\hat{j} + (4 + 2t)\hat{k}$$
$$l_2: \vec{r} = (3 + 2s)\hat{i} + (3 + 2s)\hat{j} + (2 + s)\hat{k}$$
If the co-ordinates of the point in the first octant on $$l_2$$ at a distance of $$\sqrt{17}$$ from the point of intersection of $$l$$ and $$l_1$$ are $$(a, b, c)$$, then $$18(a + b + c)$$ is equal to ______.

The direction vectors of $$l_1$$ and $$l_2$$ are $$\vec{d_1} = (1, 2, 2)$$ and $$\vec{d_2} = (2, 2, 1)$$ respectively. Since line $$l$$ passes through the origin and is perpendicular to both $$l_1$$ and $$l_2$$, its direction is $$\vec{d} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{vmatrix} = (2-4)\hat{i} - (1-4)\hat{j} + (2-4)\hat{k} = (-2, 3, -2)$$.

So line $$l$$ is $$\vec{r} = t(-2, 3, -2)$$. To find the intersection of $$l$$ and $$l_1$$, we set $$(-2t, 3t, -2t) = (3 + t_1, -1 + 2t_1, 4 + 2t_1)$$.

From the first and third components: $$-2t = 3 + t_1$$ and $$-2t = 4 + 2t_1$$. Subtracting: $$0 = -1 - t_1$$, so $$t_1 = -1$$. Then $$-2t = 3 - 1 = 2$$, giving $$t = -1$$. Verification with the second component: $$3(-1) = -3$$ and $$-1 + 2(-1) = -3$$. Confirmed.

The point of intersection is $$P_0 = (2, -3, 2)$$.

A general point on $$l_2$$ is $$(3 + 2s, 3 + 2s, 2 + s)$$. The distance from this point to $$P_0 = (2, -3, 2)$$ is $$\sqrt{17}$$:

$$(1 + 2s)^2 + (6 + 2s)^2 + s^2 = 17$$. Expanding: $$1 + 4s + 4s^2 + 36 + 24s + 4s^2 + s^2 = 17$$, which gives $$9s^2 + 28s + 37 = 17$$, so $$9s^2 + 28s + 20 = 0$$.

Using the quadratic formula: $$s = \frac{-28 \pm \sqrt{784 - 720}}{18} = \frac{-28 \pm 8}{18}$$. Thus $$s = \frac{-20}{18} = -\frac{10}{9}$$ or $$s = \frac{-36}{18} = -2$$.

For $$s = -\frac{10}{9}$$: the point is $$\left(3 - \frac{20}{9}, 3 - \frac{20}{9}, 2 - \frac{10}{9}\right) = \left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)$$, which lies in the first octant. For $$s = -2$$: the point is $$(-1, -1, 0)$$, which does not lie in the first octant.

So $$(a, b, c) = \left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)$$ and $$a + b + c = \frac{7 + 7 + 8}{9} = \frac{22}{9}$$.

Therefore, $$18(a + b + c) = 18 \times \frac{22}{9} = 44$$.