Answer
$\overrightarrow{F_{3 \space net}} = -(0.621N)\hat{i}$
Work Step by Step
Switching the sign of $Q_2$ results in reversing the direction of its force on q hence we have
$\overrightarrow{F_{3 \space net}} = (0.518 \angle - 37^o) +( 0.518\angle( 37^o - 180^o) $
$\overrightarrow{F_{3 \space net}} = (0.518 \angle - 37^o) +( 0.518\angle(-143^o) $
$\overrightarrow{F_{3 \space net}} = 0.621\angle -90^o$
$\overrightarrow{F_{3 \space net}} = -(0.621N)\hat{i}$
