Three point charges are arranged along the x-axis. Charge q1 = +2.75 µC is at the origin, and charge q2 = -6.00 µC is at x = 0.220 m. Charge q3 = -7.00 µC. Where is q3 located if the net force on q1 is 6.75 N in the −x-direction?

Answer 1

I sketched q3 between q1 and q2
look at left forces
F from 3 is left
C (6*7)/(.22-x)^2
F from 1 is also left
C (2.75*7)/(x)^2
so
C [42/(.22-x)^2 + 19.25/x^2] = 6.75
C is µ*k

Answer 2

sorry
µ^2 k

Answer 3

well, if you have two negative charges to the right of the positive, the force on the positive charge is to the right. You want it to the left, which means the q3 has to be in -x region.

netF=k( q1q2/.220^2 -q1Q3/x^2)
netF=k*10^-12(2.75*6/.220^2-2.75*7/x^2)

Then proceed to a solution. Compare this solution with yours, I do not see how you got your denominators...