Decompose your object into a triangular mesh, and inspect the intersection between the ray and the triangle.
Define your ray as passing through origin, in some direction defined by an unit vector $\hat{a}$. (This means that if your ray starts, or passes through some other point, you subtract the coordinates of that point from all triangle coordinates.)
Precalculate the unit normal vector $\hat{n}$ for each triangle beforehand. If the triangle has vertices $\vec{p}_1$, $\vec{p}_2$, and $\vec{p}_3$, then
$$\begin{aligned}
\vec{n} &= \left(\vec{p}_2-\vec{p}_1\right)\times\left(\vec{p}_3-\vec{p}_1\right) \\
\hat{n} &= \frac{\vec{n}}{\sqrt{\vec{n}\cdot\vec{n}}} \\
\end{aligned}$$
Note that $\hat{n}$ is constant, and does not depend on which point your rays might pass through.
When testing a ray, you first find out the distance $d$ where the ray intersects the plane of the triangle, and the location $\vec{p}$ of that point:
$$\begin{aligned}
d &= \hat{n} \cdot \vec{p}_1 = \hat{n} \cdot \vec{p}_2 = \hat{n} \cdot \vec{p}_3 \\
\vec{p} &= d\hat{a} \\
\end{aligned}$$
If you are not interested in the exact point where the ray intersects a given triangle, only whether the ray intersects a triangle, precalculate another three constant vectors, edge normal vectors in the plane of the triangle:
$$\begin{aligned}
\vec{e}_1 &= \hat{n} \times \left( \vec{p}_2 - \vec{p}_1 \right) \\
\vec{e}_2 &= \hat{n} \times \left( \vec{p}_3 - \vec{p}_2 \right) \\
\vec{e}_3 &= \hat{n} \times \left( \vec{p}_1 - \vec{p}_3 \right) \\
\end{aligned}$$
Note that these will point inwards. Then, knowing the point $\vec{p}$ on the plane of the triangle, the point is inside the triangle if and only if
$$\left\lbrace\begin{aligned}
\vec{e}_1 \cdot \left( \vec{p} - \vec{p}_1 \right) &\ge 0 \\
\vec{e}_2 \cdot \left( \vec{p} - \vec{p}_2 \right) &\ge 0 \\
\vec{e}_3 \cdot \left( \vec{p} - \vec{p}_3 \right) &\ge 0 \\
\end{aligned}\right.$$
If, however, you'd like to know the barycentric coordinates $(u, v)$ within the triangle, use this answer, with
$$\begin{aligned}
(x_0, y_0, z_0) &= \vec{p}_1 \\
(x_u, y_u, z_u) &= \vec{p}_2 - \vec{p}_1 \\
(x_v, y_v, z_v) &= \vec{p}_3 - \vec{p}_1 \\
\end{aligned}$$
In the $(u, v)$ planar coordinates, the three triangle vertices correspond to $(0,0)$, $(1,0)$, and $(1,1)$, respectively. Point $(u, v)$ is within the triangle if and only if $0 \le u + v \le 1$. The center of the triangle is at $(1/3, 1/3)$.
If you wish to e.g. linearly interpolate between values $c_1$, $c_2$, and $c_3$, use
$$c(u, v) = c_1 + u ( c_2 - c_1) + v (c_3 - c_1)$$
