Let the velocity of the transmitter be $\ \mathbf{v}\ $, the times at which it emits $\ T\ $ radio pulses be $\ t_1, t_2,\dots, t_T\ $, and its position at time $\ t=t_1\ $ be $\ \mathbf{x}_0\ $. All these quantities are unknown.
Let the positions of the $\ R\ $ receivers be $\ \mathbf{p}_r, r=1,2,\dots,R\ $ and the time when the $\ i^{\,\text{th}}\ $ pulse arrives at the $\ r^{\,\text{th}}\ $ receiver be $\ a_{ir}\ $. All these quantities are known.
The position of the transmitter at time $\ t\ $ will be $\ \mathbf{x}_0+\mathbf{v}\big(t-t_1\big)\ $, so the distance between the transmitter and the $\ r^{\,\text{th}}\ $ receiver at the time when it transmits the $\ i^{\,\text{th}}\ $ pulse will be $\ \big\|\big(t_i-t_1\big)\mathbf{v}+\mathbf{x}_0-\mathbf{p}_r\big\|\ $. The time the pulse will take to travel this distance is $\ c^{-1}\big\|\big(t_i-t_1\big)\mathbf{v}+\mathbf{x}_0-\mathbf{p}_r\big\|\ $, therefore it will arrive at the $\ r^{\,\text{th}}\ $ receiver at time $\ t_i+$$c^{-1}\big\|\big(t_i-t_1\big)\mathbf{v}+\mathbf{x}_0-\mathbf{p}_r\big\|\ $, which we know to be $\ a_{ir}\ $. Therefore,
\begin{align}
(1)\hspace{1em}c\big(a_{ir}-t_i\big)&=\big\|\big(t_i-t_1\big)\mathbf{v}+\mathbf{x}_0-\mathbf{p}_r\big\|
\end{align}
for $\ i=1,2,\dots,T\ $ and $\ r=1,2,\dots,R\ $. Squaring this gives
\begin{align}
(2)\hspace{1em}c^2\big(a_{ir}-t_i\big)^2&=\big\|\mathbf{v}\big\|^2\big(t_i-t_1\big)^2+2\big(t_i-t_1\big)\big\langle\mathbf{v},\mathbf{x}_0-\mathbf{p}_r\big\rangle+\big\|\mathbf{x}_0-\mathbf{p}_r\big\|^2\ .
\end{align}
In these equations, we can take $\ t_i,\ $$i=1,2,\dots,T\ $, $\big\langle\mathbf{v},\mathbf{x}_0-\mathbf{p}_r\big\rangle,\ $$\big\|\mathbf{x}_0-\mathbf{p}_r\big\|^2,\ $$r=1,2,\dots,R\ $, and $\ \big\|\mathbf{v}\big\|^2\ $ as the unknown real numbers whose values we need to recover. There are a total of $\ T+2R+1\ $ of these, and we have $\ TR\ $ equations available to determine them, so if $\ TR\ge T+2R+1\ $, then there should only be a small finite number of solutions. If $\ R\ge2\ $, this will be true if and only if $\ T\ge\frac{2R+1}{R-1}\ $. For $\ R=2\ $, $\ T=5\ $ will be enough to achieve this, while for $\ R=3\ $, $\ T=4\ $ will suffice.
Even if one of these conditions is satisfied, there's unlikely to be a convenient formula for the solutions, but it should be possible to obtain
good approximations to them by a numerical algorithm, such as the Newton-Raphson method, for instance. When the system of equations is overdetermined $\big($that is, $\ TR>T+2R+1\ $, and all the equations are $\text{used}\big)$, it will, in practice, have no exact solution anyway, because imprecision in the measurements of $\ a_{ir}\ $ means that the values we have for them are only approximations. In this case, the Newton-Raphson method will find the values of the unknown quantities that minimise the sum of squares
$$
\sum_{i=1}^T\sum_{r=1}^R\Big(c^2\big(a_{ir}-t_i\big)^2-\big\|\big(t_i-t_1\big)\mathbf{v}+\mathbf{x}_0-\mathbf{p}_r\big\|^2\Big)^2\ .
$$
Once we have determined the values of all the unknown quantities, we can obtain the speed of the transmitter by taking the square root of $\ \|\mathbf{v}\|^2\ $, and we can try to use the now known values of $\ \|\mathbf{v}\|,$$\,\big\|\mathbf{x}_0-\mathbf{p}_r\big\|^2$ and $\ \big\langle\mathbf{v},\mathbf{x}_0-\mathbf{p}_r\big\rangle\ $ to recover possible values for $\ \mathbf{x}_0\ $ and $\ \mathbf{v}\ $.
However, when $\ R=2\ $, and $\ U:\mathbb{R}^3\rightarrow\mathbb{R}^3\ $ is any rotation about the line through $\ \mathbf{p}_1\ $ and $\ \mathbf{p}_2\ $, then $\ U(\mathbf{x})=\mathbf{u}_0+\Theta\,\mathbf{x}\ $ for some orthogonal matrix $\ \Theta\ $ with $\ \mathbf{p}_r-\Theta\mathbf{p}_r=\mathbf{u}_0\ $ for $\ r=1,2\ $ $\big($because $\ U(\mathbf{p})=\mathbf{p}\ $ for all points $\ \mathbf{p}\ $ lying on the line through $\ \mathbf{p}_1\ $ and $\ \mathbf{p}_2\ \big)$, and
\begin{align}
\big\|\big(t_i-t_1\big)\mathbf{v}+\mathbf{x}_0-\mathbf{p}_r\big\|&=\big\|\Theta\big(\big(t_i-t_1\big)\mathbf{v}+\mathbf{x}_0-\mathbf{p}_r\big)\big\|\\
&=\big\|\big(t_i-t_1\big)\Theta\mathbf{v}+\Theta\mathbf{x}_0-\Theta\mathbf{p}_r\big\|\\
&=\big\|\big(t_i-t_1\big)\Theta\mathbf{v}+\Theta\mathbf{x}_0+\mathbf{u}_0-\mathbf{p}_r\big\|\ .
\end{align}
Thus, if $\ \mathbf{v}\ $,$\ \mathbf{x}_0\ $ and $\ t_i\ $, $\ i=1,2,\dots,T\ $, is any solution for the system of equations $(1)$, then so is $\ \mathbf{v}'=\Theta\,\mathbf{v}\ $, $\ \mathbf{x}_0'=\Theta\,\mathbf{x}_0+\mathbf{u}_0\ $ and $\ t_i\ $, $\ i=$$1,2,$$\dots,T\ $. Thus, with only two receivers it's impossible to recover $\ \mathbf{v}\ $ or $\ \mathbf{x}_0\ $ completely. Nevertheless, with a sufficient number of received pulses, the speed $\ \|\mathbf{v}\|\ $ of the receiver should still be uniquely determined.
When $\ R=3\ $, and $\ \mathbf{p}_1$, $\mathbf{p}_2$, $\mathbf{p}_3\ $ are not collinear, we will have known values $\ v=\|\mathbf{v}\|\ $,
\begin{align}
d_r&=\big\|\mathbf{x}_0-\mathbf{p}_r\big\|\ ,\ \text{and}\\
e_r&=\big\langle\mathbf{v},\mathbf{x}_0-\mathbf{p}_r\big\rangle
\end{align}
for $\ r=1,2,3\ $. Since $\ \mathbf{p}_3-\mathbf{p}_1\ $ and $\ \mathbf{p}_3-\mathbf{p}_2\ $ are linearly independent, we can solve the linear equations
\begin{align}
d_3^2-d_1^2-\big\|\mathbf{p}_3\big\|^2+\big\|\mathbf{p}_1\big\|^2&=2\big\langle\mathbf{p}_1-\mathbf{p}_3,\mathbf{x}_0\big\rangle\\
d_3^2-d_2^2-\big\|\mathbf{p}_3\big\|^2+\big\|\mathbf{p}_2\big\|^2&=2\big\langle\mathbf{p}_2-\mathbf{p}_3,\mathbf{x}_0\big\rangle
\end{align}
to obtain $\ \mathbf{x}_0=\mathbf{a}+\lambda\mathbf{u}\ $, where $\ \mathbf{u}\ $ is a unit vector perpendicular to $\ \mathbf{p}_3-\mathbf{p}_1\ $ and $\ \mathbf{p}_3-\mathbf{p}_2\ $, $\ \mathbf{a}\ $ a known point lying in the plane of $\ \mathbf{p}_1$, $\mathbf{p}_2$, and $\mathbf{p}_3\ $, and $\ \lambda\ $ a real parameter whose value must be either $\ \sqrt{d_1^2-\big\|\mathbf{a}-\mathbf{p}_1\big\|^2}\ $ or $\ {-}\sqrt{d_1^2-\big\|\mathbf{a}-\mathbf{p}_1\big\|^2}\ $. If $\ e_r\ $ is value recovered for $\ \big\langle\mathbf{v},\mathbf{x}_0-\mathbf{p}_r\big\rangle\ $ from the numerical solution of the system of equations $(2)$, then for each of the two possible values $\ \mathbf{a}\pm\Big(\sqrt{d_1^2-\big\|\mathbf{a}-\mathbf{p}_1\big\|^2}\Big)\mathbf{u}\ $ of $\ \mathbf{x}_0\ $, we can now solve the linear equations
$$
e_r=\big\langle\mathbf{v},\mathbf{x}_0-\mathbf{p}_r\big\rangle
$$
for $\ r=1,2,3\ $ to obtain the corresponding value of $\ \mathbf{v}\ $. The solution of either of these systems will always be the reflection of the solution of the other in the plane of $\ \mathbf{p}_1$, $\mathbf{p}_2$, and $\mathbf{p}_3\ $.
Best Answer
I will assume that the distance to the sender is much larger than the distance between any two receivers. Note that it is not possible to locate the sender exactly with only three receivers, but it is possible to estimate the direction to the sender. (In the planar example from the link there are two sensors at two different locations, so four sensors in total.) Let $v_1, v_2, v_3 \in \mathbb R^3$ be the locations of the receivers and the unit vector $n \in S^2$ the (approximate) direction to the sender. Then for two receivers $v_i, v_j$ the angle $\alpha$ between $n$ and $v_j - v_i$ satisfies $$\frac{\langle n, v_j-v_i \rangle}{\lVert v_j-v_i \rVert} = \cos(\alpha) = \frac{v \, (t_i - t_j)}{\lVert v_i - v_j \rVert}.$$ This leads to the equations $$\langle n, v_j-v_i \rangle = v \, (t_i - t_j)$$ for all pairs of indices $i, j$. Together with $\lVert n \rVert = 1$ this system has two different solutions in general, symmetric by reflection along a normal vector of the plane through $v_1, v_2, v_3$. Then write $n$ in polar coordinates.