The acceleration of the center of mass (CM) is determined by the net external force.
The net torque is the change in the angular momentum if you use the CM as the point about which to evaluate torque and angular momentum; this is true even if the CM is accelerating. This is not true using a point other than the CM for torque and angular momentum if the point is accelerating.
So the overall motion is most easily evaluated as:
(1) translational acceleration of the CM with
(2) rotational motion about the CM.
You do not need to use any "resultant force".
Addition based on your later comment specifying location of point C.
You added "5.AC=10.BC. Distance of C from 5N is twice as it is from 10N."
Then AC = 2/3 d where d is length of rod; C is d/6 to right of CM. Total torque from resultant force wrt CM is 15 * d/6 = 15/6 d = 5/2 d. This equals total torque from the actual forces: 5 * -d/2 + 10* d/2 = 5 d/2. So, C is with respect to the CM in terms of equivalent torque. That is the resultant force equivalent to the actual forces with torques evaluated wrt the CM is 15N applied at d/6 to right of CM.
You can solve the motion using either the actual forces/locations or the resultant force at location C.
Using resultant force at C, evaluate motion as translational acceleration of CM plus rotational motion about CM as other answers discuss. (If a point is not fixed or is not the CM (can be moving) net torque is not equal to the change in angular momentum. So pick CM for rotational evaluation for this problem.) CM will move upwards and rod will rotate counterclockwise wrt CM. This agrees with earlier answer by @Bob D (except for minor math error that you pointed out.)