Answer: $x_3 = -4m,\; y_2 = -3m$

Given:

Move-1: $M_1 = 5\cos 53.1^{\circ}\hat{i} + 5 \sin 53.1^{\circ} \hat{j}= 3\hat{i}+4\hat{j}$

Move-2: $M_2 = 6\hat{i} + y_2\hat{j}$

Move-3: $M_3 = x_3\hat{i} + (-3) \hat{j} = x_3\hat{i} - 3\hat{j}$

Your final placement relative to where you started: $M_f = 5.39 \cos (-21.8^{\circ})\hat{i} + 5.39 \sin(-21.8^{\circ})\hat{j} = 5\hat{i} - 2\hat{j}$

We can now equate the individual components of the moves to their final components, i.e, $\vec{M_f} = \vec{M_1} + \vec{M_2} + \vec{M_3}$

$\Rightarrow 5 = 3 + 6 + x_3 \rightarrow x_3 = - 4$

$\Rightarrow -2 = 4 + y_2 -3 \rightarrow y_2 = -3$