Решение

$\varphi_1 = \frac{kq}{r} + \frac{2kq}{2r} - \frac{3kq}{3r} = \frac{kq}{r}; \ \varphi_2 = \frac{kq}{2r} + \frac{2kq}{2r} - \frac{3kq}{3r} = \frac{kq}{2r}; \ \varphi_3 = \frac{kq}{3r} + \frac{2kq}{3r} - \frac{3kq}{3r} = 0 \ \$
$[\varphi_1 = \frac{kq}{r}; \varphi_2 = \frac{kq}{2r}; \varphi_3 = 0.]$