It's Option 2. To see this, write down the differential equations for the motion. We have Newton's Second Law as $m \dot{\vec{v}} = q (\vec{E} + \vec{v} \times \vec{B})$; assuming that both fields point in the $x$-direction ($\vec{E} = E \hat{\imath}$ and $\vec{B} = B \hat{\imath}$), the components of the Newton's Second Law are \begin{align} m \dot{v}_x &= q E\\ m \dot{v}_y &= q B v_z \\ m \dot{v}_z &= - q B v_y \end{align} You may or may not be able to solve these equations immediately. But what we can see is that the motion in the $x$-direction is completely independent of the motion in the $yz$-plane; the equation for $v_x$ doesn't include $v_y$ or $v_z$, and the equations for $v_y$ and $v_z$ don't involve $v_x$.
Moreover, the equation for $v_x$ is just what it would be for a particle in an electric field (without a magnetic field), and the equations for $v_y$ and $v_z$ are just what they would be for a charged particle in a magnetic field (without an electric field.) So the particle executes uniformly accelerated motion along the $x$-axis, and circular motion parallel to the $yz$-plane. The result is a helix whose pitch increases along the path of the particle.
Footnote: This all assumes that the charged particle is non-relativistic. If the speed of the particle becomes comparable to the speed of light, then the analysis becomes more complicated; but the basic conclusion, that the path is a helix of increasing pitch, remains essentially the same.