Answer
$U=4.77 \times 10^{-6} J \approx4.8 \times 10^{-6} J $
Work Step by Step
Electric potential Energy of 3 point charge $q_{1},q_{2},$ and $q_{3}$ is given as:
$U=\frac{1}{4\pi \epsilon_{0}}$[$\frac{q_{1}q_{2}}{r_{12}}+\frac{q_{2}q_{3}}{r_{23}}+\frac{q_{3}q_{1}}{r_{31}}]$
$=9\times 10^{9} [\frac{(2)(3)}{4}+\frac{(3)(3)}{5}+\frac{(3)(2)}{3}]\frac{10^{-18}}{10^{-2}}$
$=\frac{9\times 10^{-9}}{10^{-2}\times3}[\frac{6}{4}+\frac{9}{5}+\frac{6}{3}]$
Hence, $U=4.77 \times 10^{-6} J \approx4.8 \times 10^{-6} J$
