To find the kinetic energy of the object at the end of the displacement, we need to calculate the work done by the force $\vec{F}$ on the object during its displacement. The work done by a force is given by the dot product of the force and the displacement vectors:

$$ W = \vec{F} \cdot \vec{d} $$

Here, the force $\vec{F}$ is given as:

$$ \vec{F} = -2 \hat{i} + 3 \hat{j} \, \mathrm{N} $$

And the displacement $\vec{d}$ is given as:

$$ \vec{d} = 3 \hat{i} \, \mathrm{m} $$

The dot product of the two vectors is calculated as follows:

$$ W = (-2 \hat{i} + 3 \hat{j}) \cdot (3 \hat{i}) $$

Since the $\hat{j}$ component of the force does not contribute to the work done in the $x$-direction, it can be ignored. Thus, the dot product yields:

$$ W = -2 \cdot 3 \, \mathrm{N} \cdot \mathrm{m} = -6 \, \mathrm{J} $$

The negative sign indicates that the force does work against the direction of displacement. Now, the initial kinetic energy of the object is:

$$ K_i = 10 \, \mathrm{J} $$

The work-energy theorem relates the work done on an object to its change in kinetic energy:

$ W = K_f - K_i $

Rearranging for the final kinetic energy $K_f$ gives:

$ K_f = K_i + W $

Substituting the known values:

$$ K_f = 10 \, \mathrm{J} + (-6 \, \mathrm{J}) $$

This simplifies to:

$$ K_f = 4 \, \mathrm{J} $$

Hence, the kinetic energy of the object at the end of the displacement is Option C: $4 \mathrm{~J}$.