Let a line L_1 pass through the origin and be perpendicular to the...

The direction vectors of the given lines $L_2$ and $L_3$ are $\vec{d_2} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{d_3} = 2\hat{i} + 2\hat{j} + \hat{k}$ respectively.

Since line $L_1$ is perpendicular to both $L_2$ and $L_3$, its direction vector $\vec{d_1}$ is given by the cross product of $\vec{d_2}$ and $\vec{d_3}$:
$\vec{d_1} = \vec{d_2} \times \vec{d_3} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{vmatrix} = -2\hat{i} + 3\hat{j} - 2\hat{k}$

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

Let the point of intersection of $L_1$ and $L_2$ be $P$. The coordinates of $P$ can be written as $(-2\lambda, 3\lambda, -2\lambda)$ from $L_1$ and $(3+t, 2t-1, 2t+4)$ from $L_2$. Equating the respective coordinates:
$-2\lambda = 3 + t$
$3\lambda = 2t - 1$
$-2\lambda = 2t + 4$

From the first and third equations, we get:
$3 + t = 2t + 4 \Rightarrow t = -1$

Substituting $t = -1$ into the first equation gives $-2\lambda = 2 \Rightarrow \lambda = -1$. This also satisfies the second equation. Thus, the point of intersection is $P(2, -3, 2)$.

Let the point on $L_3$ be $Q(a, b, c) = (3+2s, 3+2s, 2+s)$. The distance between $P$ and $Q$ is given as $\sqrt{17}$.
$PQ^2 = 17$
$(3+2s - 2)^2 + (3+2s - (-3))^2 + (2+s - 2)^2 = 17$
$(2s + 1)^2 + (2s + 6)^2 + s^2 = 17$
$4s^2 + 4s + 1 + 4s^2 + 24s + 36 + s^2 = 17$
$9s^2 + 28s + 37 = 17$
$9s^2 + 28s + 20 = 0$
$9s^2 + 18s + 10s + 20 = 0$
$(9s + 10)(s + 2) = 0$

This gives $s = -2$ or $s = -\dfrac{10}{9}$.

Since $a = 3 + 2s$ must be an integer ($a \in \mathbb{Z}$), we must choose $s = -2$.

Substituting $s = -2$ into the coordinates of $Q$:
$a = 3 + 2(-2) = -1$
$b = 3 + 2(-2) = -1$
$c = 2 + (-2) = 0$

Therefore, $(a, b, c) = (-1, -1, 0)$.

We need to find the value of $(a+b+c)^2$:
$(a+b+c)^2 = (-1 - 1 + 0)^2 = (-2)^2 = 4$

Answer: $4$