1. Given relation for three forces

If three vectors are represented by the three sides of a triangle taken in order, then their vector sum is zero:

$\overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3}=0$

or equivalently,

$\overrightarrow{F_1}+\overrightarrow{F_2}=-\overrightarrow{F_3}$

This is the triangle law of forces.

2. Check Statement I

Statement I says:

If three forces $\overrightarrow{F_1}, \overrightarrow{F_2}, \overrightarrow{F_3}$ are represented by three sides of a triangle and $\overrightarrow{F_1}+\overrightarrow{F_2}=-\overrightarrow{F_3},$ then these three forces are concurrent forces and satisfy the condition for equilibrium.

Since the three forces form a closed triangle, their resultant is zero:

$\sum \vec F = 0$

So the condition for translational equilibrium is satisfied.

Also, for three forces to keep a rigid body in equilibrium, if they are non-parallel, their lines of action must be concurrent. This is the standard condition for equilibrium of three coplanar forces.

Hence Statement I is true.

3. Check Statement II

Statement II says:

A triangle made up of three forces $\overrightarrow{F_1}, \overrightarrow{F_2}$ and $\overrightarrow{F_3}$ as its sides taken in the same order, satisfy the condition for translatory equilibrium.

If three forces taken in order form a triangle, then:

$\overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3}=0$

This is exactly the condition for translatory equilibrium.

So Statement II is also true.

4. Final choice

Both Statement I and Statement II are true.

Therefore, the correct option is:

$\boxed{D}$