Answer
$$V=\frac{1}{4\pi\epsilon_o}(\frac{3Q}{\mathcal{l}}+\frac{Q}{\sqrt2\mathcal{l}}+\frac{-2Q}{\mathcal{l}})$$
Work Step by Step
Use equation 17–5 to find the potential of a single point charge. The potential at the corner is the sum of the potentials due to each of the other 3 charges.
$$V=\frac{1}{4\pi\epsilon_o}(\frac{Q_1}{r_1}+\frac{Q_2}{r_2}+\frac{Q_3}{r_3})$$
$$V=\frac{1}{4\pi\epsilon_o}(\frac{3Q}{\mathcal{l}}+\frac{Q}{\sqrt2\mathcal{l}}+\frac{-2Q}{\mathcal{l}})$$
