A force F acts on an object that moves to the right on a rough, horizontal surface. The velocity of the object decreases. The directions of the resultant force and the acceleration are:

The block has a mass $m = 3 \text{ kg}$.
There are two applied forces:

• A horizontal force $F_1 = 12 \text{ N}$ to the left.
• A force $F_2 = 30 \text{ N}$ at an angle of $60^\circ$ above the horizontal, pointing to the right.

We need to resolve $F_2$ into its horizontal ($F_{2x}$) and vertical ($F_{2y}$) components:

$$F_{2x} = F_2 \cos(60^\circ) = 30 \text{ N} \times 0.5 = 15 \text{ N} \text{ (to the right)}$$ $$F_{2y} = F_2 \sin(60^\circ) = 30 \text{ N} \times \frac{\sqrt{3}}{2} \approx 30 \text{ N} \times 0.866 = 25.98 \text{ N} \text{ (upwards)}$$

Let's consider the right direction as positive.

$$\sum F_x = F_{2x} - F_1 = 15 \text{ N} - 12 \text{ N} = 3 \text{ N}$$

The net horizontal force is $3 \text{ N}$ to the right. Since this is not zero, the block will accelerate horizontally.

The gravitational force acting on the block is $F_g = mg$.

$$F_g = 3 \text{ kg} \times 9.8 \text{ m/s}^2 = 29.4 \text{ N} \text{ (downwards)}$$

The upward component of the applied force is $F_{2y} = 25.98 \text{ N}$.
To lift off the floor, the total upward force must be greater than the gravitational force.
In this case, $F_{2y} < F_g$ ($25.98 \text{ N} < 29.4 \text{ N}$).
Therefore, the block will not lift off the floor. There will be a normal force from the surface supporting the block.

Since there is a net horizontal force of $3 \text{ N}$ to the right, and the block does not lift off the floor, the block will accelerate to the right.

The final answer is $\boxed{\text{C. accelerate to the right.}}$