The correct option is A may pass though a continuous charge distribution
A gaussian surface can well pass through a continuous charge distribution (For example: a volume charge distribution) as the field everywhere (even on the gaussian surface) is defined for such a distribution. The simplest example is that of uniform solid non-conducting sphere with uniform volume charge. The field inside grows linearly with radial distance and falls off as the square of radial distance outside. So, any gaussian surface drawn in the interior of solid sphere passes through a continuous charge distribution.