Let, as a model, we use a camera point $K$ and a camera screen $k$, where $k$ is a plane, and let the orthogonal projection $O$ of point $K$ on the plane $k$ be the origin of the two dimensional coordinate system $Oe_1e_2$ on the screen $k$. Attach a three dimensional coordinate system $Ke_1e_2e_3$ to the point $K$ where $e_1$ and $e_2$ are the unit vectors of $Oe_1e_2$ and let $e_3$ be the unit vector orthogonal to $k$ and pointing from $K$ to $k$. Denote by $\nu$ the distance from $K$ to $k$. We may or may not know it.
At first let $A_0B_0C_0$ be an equilateral triangle in a plane $p$ parallel to $k$. Let $G_0 \in p$ be the center of $A_0B_0C_0$ so that $G_0$ lies on the $e_3$ axis, i.e. $G_0$ is the orthogonal projection of $K$ onto $p$. Then we project $A_0B_0C_0$ onto the plane $k$ by drawing the lines $KA_0, \, KB_0, \, KC_0$ and possibly $KG_0$ and denoting by $A, B, C$ and $O$ their intersection points with $k$. Thus we obtain the projected equilateral triangle $ABC$ with center $O$.
Next we move the camera $K$ together with the screen $k$ to a different position and again we project the triangle $A_0B_0C_0$ with center $G_0$ onto the plane $k$ (which is now positioned differently with respect to $p$ and is this possibly not parallel to it) the same way we did before. We draw the lines between point $K$ and the points $A_0, B_0, C_0$ and $G_0$ and look at their respective intersection points $A, B, C$ and $G$ with plane $k$. The resulting triangle $ABC$ is most likely not equilateral.
Think of the coordinate system $Ke_1e_2e_3$, which moved together with the camera and the screen, and recall that $\nu$ is the distance between $K$ and $k$, the orthogonal projection of $K$ onto $k$ is $O$, and $Oe_1e_2$ is the coordinate system on the screen $k$ as before.
If we take a plane $q$ parallel to $p$ passing through point say $C$, then the intersection points
$$A_q = KA \cap q, \,\, B_q = KB \cap q, \,\, C_q = KC \cap q \, \text{ and } \, G_q = KG \cap q$$
form an equilateral triangle $A_qB_qC$ with center $G_q$ whose edges are parallel to the edges of $A_0B_0C_0$ (in other words triangle $A_qB_qC$ with center $G_q$ is homothetic image of triangle $A_0B_0C_0$ with center $G_0$ with respect $K$ )
Now, you know the locations of the points $A, \, B, \, C$ and say $G$ with respect to the screen's $k$ coordinate system $Oe_1e_2$ which means you know the vectors $$\vec{OA}, \, \vec{OB}, \, \vec{OC}, \,\, \text{ and } \,\, \vec{OG}$$ so thus, we can form the vectors
\begin{align*}
\vec{KA} &= \vec{KO} + \vec{OA} = \vec{OA} + \vec{KO} = \vec{OA} + \nu \, e_3\\
\vec{KB} &= \vec{KO} + \vec{OB} = \vec{OB} + \vec{KO} = \vec{OB} + \nu \, e_3\\
\vec{KC} &= \vec{KO} + \vec{OC} = \vec{OC} + \vec{KO} = \vec{OC} + \nu \, e_3\\
\vec{KG} &= \vec{KO} + \vec{OG} = \vec{OG} + \vec{KO} = \vec{OG} + \nu \, e_3\\
\end{align*}
Thus, there exists constants $\lambda, \, \mu$ and $\nu$, if you do not know $\nu$
\begin{align*}
\vec{KA_q} &= \lambda \, \vec{KA}= \lambda \, \vec{OA} + \lambda \, \nu \, e_3\\
\vec{KB_q} &= \mu \, \vec{KB}= \mu \, \vec{OA} + \mu \, \nu \, e_3\\
\vec{KG_q} &= \frac{1}{3} \big(\lambda \, \vec{KA} + \mu \, \vec{KB} + \vec{KC}\big)\\ &= \frac{1}{3} \big( \lambda \, \vec{OA} + \mu \, \vec{OA} + \vec{OC} + (\lambda + \mu + 1) \, \nu \, e_3 \big)
\end{align*}
The restrictions that lead to the equations for the unknowns $\lambda, \, \mu$ and possibly $\nu$ are
\begin{align*}
|\vec{C_qA_q}| &= |\vec{A_qB_q}|\\
|\vec{C_qB_q}| &= |\vec{A_qB_q}|\\
\vec{KG_q} & \, |\, | \, \vec{KG}\\
\end{align*} where the first two say that the edges of equilateral triangle $A_qB_qC$ are equal while the last one represents the fact that the points $K, \, G_q $ and $G$ are collinear. As equations, these three conditions are
\begin{align*}
|\lambda \, \vec{KA} - \vec{KC}|^2 &= |\lambda \, \vec{KA} - \mu \, \vec{KB}|^2 \\
|\lambda \, \vec{KA} - \vec{KC}|^2 &= |\lambda \, \vec{KA} - \mu \, \vec{KB}|^2 \\
|\vec{KG_q}\times \vec{KG}|^2 &= 0\\
\end{align*}
which written in terms of $\lambda, \, \mu, \, \nu$ unknown variables turn into
\begin{align*}
|\lambda \, \vec{OA} - \vec{OC} + (\lambda \, \nu \, - \nu) \,
e_3 |^2 &= |\lambda \, \vec{OA} - \mu \, \vec{OB} + (\lambda \,
\nu \, - \mu \, \nu) \,
e_3 |^2 \\
|\mu \, \vec{OB} - \vec{OC} + (\mu \, \nu \, - \nu) \, e_3 |^2 &=
|\lambda \, \vec{OA} - \mu \, \vec{OB} + (\lambda \, \nu \, - \mu
\, \nu) \,
e_3 |^2 \\
|\big(\frac{1}{3} \big( \lambda \, \vec{OA} + \mu \,
\vec{OA} + \vec{OC} + (\lambda + \mu + 1) \, \nu \, e_3 \big)\big)
\times \big(\vec{OG} + \nu \, e_3 \big) |^2 &= 0\\
\end{align*}
If you know $\nu$ then only the first two equations are relevant
\begin{align*}
|\lambda \, \vec{OA} - \vec{OC} + (\lambda \, \nu \, - \nu) \,
e_3 |^2 &= |\lambda \, \vec{OA} - \mu \, \vec{OB} + (\lambda \,
\nu \, - \mu \, \nu) \,
e_3 |^2 \\
|\mu \, \vec{OB} - \vec{OC} + (\mu \, \nu \, - \nu) \, e_3 |^2 &=
|\lambda \, \vec{OA} - \mu \, \vec{OB} + (\lambda \, \nu \, - \mu
\, \nu) \,
e_3 |^2
\end{align*}
All these systems of equations may have several solutions though.
I haven't checked for typos and mistakes and I do not guarantee this is the best solution.
