Calculate the net force vector F.

The particle is acted on by three constant forces. Let's sum them:
$\vec{F_1} = 2\vec{i} + \vec{j} - \vec{k}$
$\vec{F_2} = \vec{k} - \vec{j}$ (which can be written as $0\vec{i} - \vec{j} + \vec{k}$)
$\vec{F_3} = 3\vec{i} + \vec{j} + 5\vec{k}$

The net force $\vec{F}$ is the sum of these forces:

$$\vec{F} = \vec{F_1} + \vec{F_2} + \vec{F_3}$$ $$\vec{F} = (2\vec{i} + \vec{j} - \vec{k}) + (0\vec{i} - \vec{j} + \vec{k}) + (3\vec{i} + \vec{j} + 5\vec{k})$$ $$\vec{F} = (2+0+3)\vec{i} + (1-1+1)\vec{j} + (-1+1+5)\vec{k}$$ $$\vec{F} = 5\vec{i} + 1\vec{j} + 5\vec{k}$$ $$\vec{F} = 5\vec{i} + \vec{j} + 5\vec{k}$$

The particle is displaced from point $\vec{r_1} = \vec{i} + 2\vec{j} + 3\vec{k}$ to point $\vec{r_2} = 6\vec{i} + 3\vec{j} + \vec{k}$.
The displacement vector $\vec{d}$ is given by $\vec{r_2} - \vec{r_1}$:

$$\vec{d} = (6\vec{i} + 3\vec{j} + \vec{k}) - (\vec{i} + 2\vec{j} + 3\vec{k})$$ $$\vec{d} = (6-1)\vec{i} + (3-2)\vec{j} + (1-3)\vec{k}$$ $$\vec{d} = 5\vec{i} + 1\vec{j} - 2\vec{k}$$ $$\vec{d} = 5\vec{i} + \vec{j} - 2\vec{k}$$

The work done by a constant force is given by the dot product of the force vector and the displacement vector:

$$W = \vec{F} \cdot \vec{d}$$ $$W = (5\vec{i} + \vec{j} + 5\vec{k}) \cdot (5\vec{i} + \vec{j} - 2\vec{k})$$ $$W = (5)(5) + (1)(1) + (5)(-2)$$ $$W = 25 + 1 - 10$$ $$W = 26 - 10$$ $$W = 16$$

The work done is 16 units (assuming standard units like Joules if force is in Newtons and displacement in meters).

The final answer is 16.