Where should a third charge of +7 micro Coulombs be placed so that the electric field at x=0 cm is zero?

First draw what the question is asking.
make sure your units are accurate. since the answer will be in cm. there is no need to change the
-5cm or the 3cm to meters.
Notice that +2micro=+2E-6Coulombs and that -10micro=-10E-6Coulombs as well.
The Information so far you have is
for q1 (or charge 1) you have +2E-6C, distance(R1)= -5cm from the origin
for q2 (or charge 2) you have -10E-6C, distance(R2)= 3cm from the origin
for q3 +7E-6C, distance(x) from the origin
Notice that q1 is positive, therefore it follows that it's electric field will act on the origin to the right.
Notice that q2 is negative, therefore it follows that it's electric field will act on the origin to the right.
The equation for the electric field follows the law of superposition.
The equation for the electric force follows the law of superposition.
Force follows the the law of superposition.
Thus all are summative.

Calculate individually, the nature of the electric fields on the origin first. do this for q1 then q2.
E1= absolute value of (charge at origin)X(q1)X(coulomb konstant (K))/ (R1^2)/ (charge at origin)
The charge at the origin will cancel. leaving (q1)(K)/(R1^2)= E1
E2 = the same exact procedure.

At this point E1+E2 will give E3 (vectorially only) normally you break E1 and E2 into components.
Here E3= the electric field at the origin.
Things to remember: As a vector E1 points toward the origin, and E2 points away from the origin, this is due to there electric fields. further, no vector is in the y axis, thus there is no y component.

thus E1 has a direction of 0degrees, and E2 has a direction of 0 degrees with respect to the origin.
calculate E1(cos(0)) to get x component.
calculate E2(cos(0)) to get x component.
Add the components. This is the sum of the electric fields or E3.
The question wants this field to be zero. In other words the +7E-6C charge will be placed on the x axis some distance to block the Electric field there. The charge itself sets up an electric field so will affect the origin.
now q3= +7E-6C at some distance x from the origin
calculate its electric field magnitude:  (q3)(K)/x^2 = E4. However E4 needs to equal E3 in order that the field at the origin is zero.
(q3)(K)/x^2= E3.. Now solve for x. this is the distance the charge must be away from the origin so that the field there results in zero by the law of superposition. you will get a positive number, since all E values are absolute.

you can chck this by adding all of the fields together. if you get zero you are correct. in other words after you find x, plug that back in to (q3)(K)/x^2. you will get E3. but remember that the vector is in the opposite direction since the charge is positive, toward the origin, so they cancel. E3+-E3 = 0.