Let $a, b, c \in \mathbb{C}$ be the affixes of the vertices of triangle $ABC$. For $t \ge 0$, we have the following expressions for the affixes of the points $M(t)$, $N(t)$, and $P(t)$:
$$\left\{ \begin{array}{l} m(t) = b \cdot (1 - v_1 \cdot t) + c \cdot v_1 \cdot t; \\ n(t) = c \cdot (1 - v_2 \cdot t) + a \cdot v_2 \cdot t; \\ p(t) = a \cdot (1 - v_3 \cdot t) + b \cdot v_3 \cdot t. \end{array} \right.$$

The condition that the triangle $MNP$ is equilateral at some time $t \ge 0$ can be written as:
$$m(t) + \varepsilon n(t) + \bar{\varepsilon} p(t) = 0,$$
where $\epsilon \in \left\{ \frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2} \right\}$.
This relationship is equivalent to:
$$t(-bv_1 + cv_1 - \epsilon cv_2 + \epsilon av_2 - \bar{\epsilon}av_3 + \bar{\epsilon}bv_3) + b + \epsilon c + \bar{\epsilon}a = 0, \quad (*)$$
for any $t \ge 0$ for which the triangle $MNP$ is equilateral.

a) Since the relation $(*)$ holds for three distinct values $t_1, t_2, t_3 \ge 0$, according to the pigeonhole principle, there exists $\epsilon \in \left\{ \frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2} \right\}$ for which the relation $(*)$ is satisfied at two distinct moments $t_i, t_j \ge 0$, $1 \le i < j \le 3$, from which it follows that:
$$\begin{cases} (b-c)v_1 + \epsilon(c-a)v_2 + \bar{\epsilon}(a-b)v_3 = 0; \\ b + \epsilon c + \bar{\epsilon} a = 0. \end{cases}$$
The second relation from above is equivalent to having an equilateral triangle $ABC$. On the other hand, the first relationship allows us to translate the triangle $ABC$ such that its circumcenter has the affix $0$. Since $ABC$ is equilateral, we have $b = \epsilon a$ and $c = \bar{\epsilon} a$ or $b = \bar{\epsilon} a$ and $c = \epsilon a$, without loss of generality, we consider the first case and obtain:
$$(\epsilon a - \bar{\epsilon} a)v_1 + \epsilon(\bar{\epsilon} a - a)v_2 + \bar{\epsilon}(a - \epsilon a)v_3 = 0 \Rightarrow (\epsilon - \epsilon^2)v_1 + (1 - \epsilon)v_2 + (\epsilon^2 - 1)v_3 = 0,$$
$\text{and}$ by dividing with } 1 - $\epsilon \neq 0 \text{}$ and taking into account that } v_1, v_2, v_3 $\in \mathbb{R}, \text{}$ we obtain:}
$$\epsilon v_1 + v_2 - v_3 - \epsilon v_3 = 0 \Rightarrow v_1 = v_2 = v_3.$$

b) If $v_1 = v_2 = v_3 = v$, relation $(*)$ can be written as:
$$t \cdot v \cdot ((c - b) + \epsilon(a - c) + \bar{\epsilon}(b - a)) + b + \epsilon c + \bar{\epsilon} a = 0,$$
which is invariant under translations, rotations, and homotheties, so we can fix $a = 1$ and $c = \bar{\epsilon}$ and obtain:
$$(b - \epsilon)(t \cdot v \cdot (\bar{\epsilon} - 1) + 1) = 0,$$
and, since the second term cannot be zero, we have $b = \epsilon$, from which we obtain that triangle $ABC$ is equilateral, and thus concludes the problem.