Three charged particles are placed at the corners of an equilateral triangle of side d = 1.00 m.The charges are Q1 = +4.0 µC, Q2 = -7.0 µC, and...

The force of attraction between Q1and Q2 is
Q1*Q2/[(4peo)d^2] = 9*10^9*4*7*10^-(12) /1^2 N=0.252N

The
force of attraction between Q1 and Q3 = 9*10^8(4*6/1^2) N = 0.192
N.

So the net force of 0.252N and 0.192N with 60 degree
between is sqrt(0.252^2+0.192^2+ 2*0.252*0*192cos60) =
0.3856839N

The angle between 0.252N and the resultant
0.3857N is

cos inverse
(0.38565839^2+0.252^2-0.192^2)/(2*0.38565839*0.252) = 27.08
degree.

Similarly you can proceed to find the magnitude and
direction of the resultant force exerted on other two charges.