This question depends on the specific charge values and figure (triangle ABC with charges at each vertex), which were not preserved in the extracted text. General method: by symmetry, if the two charges at B and C are equal, their individual field contributions at the midpoint M (equidistant from both) have components along BC that cancel and components along the median AM that add — so their net field lies along the line AM. Adding the field due to the charge at A (directed along AM or MA depending on its sign) then gives the total field, which will be directed along AM (or MA) unless the contributions exactly cancel (giving zero). Please refer to the original figure/charge values to select the correct final option.