Three point charges q_1 = 1\ , q_2 = -2\ , and q_3 = 3\ are placed...

The electric potential at the origin is given by the superposition principle:

$V = \dfrac{1}{4\pi\epsilon_0} \left( \dfrac{q_1}{r_1} + \dfrac{q_2}{r_2} + \dfrac{q_3}{r_3} \right)<br><br>The distances$ of the charges from the origin are:

r_1 = $\sqrt{1^2 + 0^2 + 0^2} =$1\ $\text{m}<br><br>$r_2 = $\sqrt{0^2 + 2^2 + 0^2} =$2\ $\text{m}<br><br>$r_3 = $\sqrt{0^2 + 0^2 + 3^2} =$3\ $\text{m}<br><br>Substituting$ the values of charges and distances:

V = 9 $\times$10^9 \left( \dfrac{1 \times 10^{-6}}{1} + \dfrac{-2 \times 10^{-6}}{2} + \dfrac{3 \times 10^{-6}}{3} \right)$<br><br>$V = 9 \times 10^9 \times 10^{-6} (1 - 1 + 1)$<br><br>$V = 9 \times 10^3\ \text{V}$<br><br>Answer:$9 \times 10^3\ \text{V}