1. Since the particle remains at rest under the three forces, their vector sum must be zero:

$\vec F_1 + \vec F_2 + \vec F_3 = 0$

So,

$\vec F_1 = -\left(\vec F_2 + \vec F_3\right)$

This means the magnitude of $F_1$ must equal the resultant of $F_2$ and $F_3$.

2. Given that $F_2$ and $F_3$ are perpendicular,

$R = \sqrt{F_2^2 + F_3^2} = \sqrt{8^2 + 6^2} = \sqrt{64+36} = \sqrt{100} = 10\,\text{N}$

This matches $F_1 = 10\,\text{N}$, so equilibrium is consistent.

3. If $F_1$ is removed, only $F_2$ and $F_3$ act on the particle. Their resultant force is:

$F_{\text{net}} = \sqrt{8^2 + 6^2} = 10\,\text{N}$

4. Using Newton's second law,

$a = \frac{F_{\text{net}}}{m} = \frac{10}{5} = 2\,\text{m/s}^2$

5. Hence the acceleration of the particle is:

$\boxed{2\,\text{m/s}^2}$

So the correct option is C.