Answer
Rightward, $\ \ 63\ \mathrm{N}$
Work Step by Step
Electric charges exert forces on one another along the line connecting them: like charges repel, opposite charges attract.
The magnitude of the force between two point charges, $q_{1}$ and $q_{2}$, separated by a distance $r$ is
$F =k\displaystyle \frac{|q_{1}||q_{2}|}{r^{2}},\qquad (19- 5)$
The constant $k$ in this expression is $k=8.99\times 10^{9}\mathrm{N}\cdot \mathrm{m}^{2}/\mathrm{C}^{2} ,\qquad (19-6)$
The electric force on one charge due to two or more other charges is the vector sum of each individual force.
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Taking rightward to be the +x direction, $\mathrm{q}_{2}$ acts with attractive force and $\mathrm{q}_{3}$ with a repelling force on $\mathrm{q}_{1}.$
$\displaystyle \mathrm{F}=+k\frac{q_{1}q_{2}}{d^{2}}-k\frac{q_{1}q_{3}}{(2d)^{2}}$
$=\displaystyle \frac{k\mathrm{q}_{1}}{d^{2}}(q_{2}-\frac{q_{3}}{4})$
$=\displaystyle \frac{k\mathrm{q}}{d^{2}}(2.0\mathrm{q}-\frac{3.0\mathrm{q}}{4})$
$=\displaystyle \frac{k\mathrm{q}}{d^{2}}(\frac{5}{4}\mathrm{q})$
$=\displaystyle \frac{5}{4}\cdot\frac{(8.99\times 10^{9}\mathrm{N}\cdot \mathrm{m}^{2}/\mathrm{C}^{2})(12\times 10^{-6}\mathrm{C})^{2}}{(0.16\mathrm{m})^{2}}$
$=+63\mathrm{N}$
