= Solution
In a homogeneous, time-independent medium, $\partial\Omega/\partial x_i=0$ and $\partial\Omega/\partial t=0$. Therefore $\mathbf k$ and $\omega$ are constant along each ray. The <group velocity> $\mathbf c_g=\nabla_{\mathbf k}\Omega$ is consequently constant, and
$$
\mathbf x(t)=\mathbf x(0)+t\mathbf c_g,
$$
so every ray is a straight line.
For $\omega=0$, let $s=|\mathbf x|$ denote distance along a ray from the origin and let
$$
\widehat{\mathbf c}_g=\frac{\mathbf c_g}{|\mathbf c_g|}.
$$
The phase equation gives
$$
\frac{d\Theta}{ds}
=\frac1{|\mathbf c_g|}\frac{d\Theta}{dt}
=\widehat{\mathbf c}_g\mathbin{\cdot}\mathbf k.
$$
Both vectors on the right are constant along the ray, so
$$
\boxed{
\Theta(\mathbf x)=\Theta(\mathbf0)
+\widehat{\mathbf c}_g\mathbin{\cdot}\mathbf k\,|\mathbf x|.}
$$
The final formula in the question denotes the physical phase $\Theta=\theta/\varepsilon$ simply by $\theta$, giving its stated expression.
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