Answer
The x-coordinate of particle 3 is $~~x = -13.7~cm$
Work Step by Step
If the net electrostatic force on particle 3 due to the other two particles is zero, then the force vectors due to the other two particles must point in opposite directions. This is only possible if particle 3 is located on the x-axis.
If particle 3 is located between particle 1 and particle 2, then the forces on particle 3 due to the other two particles are in the same direction. Then the net force can not be zero.
If particle 3 is located to the right of particle 2, then the magnitude of the force on particle 3 due to particle 2 is greater than the magnitude of the force on particle 3 due to particle 1. Then the net force can not be zero.
Therefore, particle 3 must be located on the x-axis to the left of particle 1.
We can find $r_1$, the distance between particle 1 and particle 3:
$\frac{\vert q_1 \vert~\vert~q_3 \vert}{4\pi~\epsilon_0~r_1^2} = \frac{\vert q_2 \vert~\vert~q_3 \vert}{4\pi~\epsilon_0~r_2^2}$
$\frac{\vert q_1 \vert}{r_1^2} = \frac{\vert q_2 \vert}{r_2^2}$
$\frac{\vert q_1 \vert}{r_1^2} = \frac{\vert q_2 \vert}{(r_1+10.0)^2}$
$\vert q_1 \vert~(r_1+10.0)^2 = \vert q_2 \vert r_1^2$
$q_1~(r_1^2+20.0~r_1+100) = 3q_1 r_1^2$
$r_1^2+20.0~r_1+100 = 3r_1^2$
$2r_1^2-20.0~r_1-100 = 0$
We can use the quadratic formula:
$r_1 = \frac{20.0\pm \sqrt{(-20.0)^2-(4)(2)(-100)}}{(2)(2)}$
$r_1 = \frac{20.0\pm \sqrt{1200}}{4}$
$r_1 = -3.66~cm, 13.7~cm$
Since the distance $r_1$ must be positive, the solution is $r_1 = 13.7~cm$
Note that particle 3 must be located on the x-axis to the left of particle 1.
The x-coordinate of particle 3 is $~~x = -13.7~cm$
