Let a line $$L_1$$ pass through the origin and be perpendicular to the lines
$$L_2 : \vec{r} = (3+t)\hat{i} + (2t-1)\hat{j} + (2t+4)\hat{k}$$ and
$$L_3 : \vec{r} = (3+2s)\hat{i} + (3+2s)\hat{j} + (2+s)\hat{k}$$, $$t, s \in \mathbf{R}$$.
If $$(a, b, c)$$, $$a \in \mathbb{Z}$$, is the point on $$L_3$$ at a distance of $$\sqrt{17}$$ from the point of intersection of $$L_1$$ and $$L_2$$, then $$(a + b + c)^2$$ is equal to __________.

To solve this, we need to find the direction of $$L_1$$, the intersection point of $$L_1$$ and $$L_2$$, and finally the point on $$L_3$$.

1. Direction of $$L_1$$

$$L_1$$ is perpendicular to $$L_2$$ (direction $$\vec{v_2} = \hat{i} + 2\hat{j} + 2\hat{k}$$) and $$L_3$$ (direction $$\vec{v_3} = 2\hat{i} + 2\hat{j} + \hat{k}$$).

The direction of $$L_1$$ is $$\vec{v_1} = \vec{v_2} \times \vec{v_3}$$:

$$\vec{v_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{vmatrix} = \hat{i}(2-4) - \hat{j}(1-4) + \hat{k}(2-4) = -2\hat{i} + 3\hat{j} - 2\hat{k}$$

Since $$L_1$$ passes through the origin, its equation is $$\vec{r} = \lambda(-2\hat{i} + 3\hat{j} - 2\hat{k})$$.

2. Intersection of $$L_1$$ and $$L_2$$

Equate $$L_1$$ and $$L_2$$:

$$\lambda(-2) = 3 + t \implies t = -2\lambda - 3$$

$$\lambda(3) = 2t - 1 \implies 3\lambda = 2(-2\lambda - 3) - 1 \implies 3\lambda = -4\lambda - 7 \implies 7\lambda = -7 \implies \lambda = -1$$

Substituting $$\lambda = -1$$ into $$L_1$$, the intersection point $$P$$ is $$(2, -3, 2)$$.

3. Find Point $$(a, b, c)$$ on $$L_3$$

Any point on $$L_3$$ is $$Q = (3+2s, 3+2s, 2+s)$$. The distance $$PQ = \sqrt{17}$$:

$$(3+2s - 2)^2 + (3+2s - (-3))^2 + (2+s - 2)^2 = 17$$

$$(2s+1)^2 + (2s+6)^2 + s^2 = 17$$

$$4s^2 + 4s + 1 + 4s^2 + 24s + 36 + s^2 = 17$$

$$9s^2 + 28s + 20 = 0$$

Using the quadratic formula: $$s = \frac{-28 \pm \sqrt{784 - 720}}{18} = \frac{-28 \pm 8}{18}$$

$$s = -\frac{20}{18} = -\frac{10}{9}$$ or $$s = -\frac{36}{18} = -2$$

We are given $$a \in \mathbb{Z}$$ (an integer).

If $$s = -2$$, then $$a = 3 + 2(-2) = -1$$ (integer).

The point $$(a, b, c)$$ is:

$$a = 3 - 4 = -1$$

$$b = 3 - 4 = -1$$

$$c = 2 - 2 = 0$$

4. Final Calculation

$$(a + b + c)^2 = (-1 - 1 + 0)^2 = (-2)^2 = 4$$

Correct Answer: 4