Answer: Given,

To find the resultant force

$\sum F_x= 150-250\cos{60}-300\sin{60} \\=-125N \\\text{and} \sum{F_y}=250\sin{60}-300\cos{30} = 43.30N $

$\therefore \text{ Resultant Force } R = \sqrt{ \left( \sum{F_x} \right)^2+\left( \sum{F_y} \right)^2} \\= \sqrt{ \left(-125\right)^2 +\left(43.3\right)^2} = 132.29N$

and Direction 

$\theta = \tan^{-1}\left( \dfrac{\sum{F_y}}{\sum{F_x}} \right) \\= \tan^{-1}\left( \dfrac{-43.3}{-125} \right) = 198.98^0$

ie. resultant force 132.29N acts at an angle of 198.980 with horizontal (in anticlockwise direction).

Also, Equillibrant force is a force equal to, but opposite of, the resultant sum of vector forces.

Hence, equillibrant force for the above system will be force with magnitude 132.29N acting at an angle of 18.980 

(angle obtained by subtracting 1800 as resultant force and equillibrant forces are opposite to each other) with horiziontal.