Answer
$\vec F_3=(5.2915N)\hat x-(6.0008N)\hat y$
Work Step by Step
We know that
$\vec F=\vec F_1+\vec F_2+\vec F_3$
$\implies m\vec a=\vec F_1+\vec F_2+\vec F_3$
$\implies (5.95Kg)[(1.17m/s^2)\hat x+(-0.664m/s^2)\hat y]=(3.22N)\hat x+(-1.55N)\hat x+(2.05N)\hat y+F_{3x}\hat x+F_{3y}\hat y$
This simplifies to:
$F_{3x}\hat x+F_{3y}\hat y=(5.2915N)\hat x-(6.0008N)\hat y$
$\implies \vec F_3=(5.2915N)\hat x-(6.0008N)\hat y$
