Write $a = a_\parallel + a_\perp$ and $c = c_\parallel + c_\perp$, where $a_\parallel, c_\parallel$ are the projections onto $b$ and $a_\perp, c_\perp$ are perpendicular to $b$.
We are given $\langle a,b \rangle = |a||b|\cos\theta_1 = \frac12|a||b|$, but we also have $\langle a, b \rangle = |a_\parallel||b|$, so we know that $|a_\parallel| = \frac12 |a|$. Similarly, $|c_\parallel| = \frac12|c|$.
Because $|a|^2 = |a_\parallel|^2 + |a_\perp|^2$ (which we can derive by expanding $\langle a_\parallel + a_\perp, a_\parallel + a_\perp\rangle$) we can also deduce that $|a_\perp| = \frac{\sqrt3}2 |a|$. Similarly, $|c_\perp| = \frac{\sqrt3}2
|c|$. Therefore
\begin{align}
\langle a,c\rangle &= \langle a_\parallel + a_\perp, c_\parallel + c_\perp\rangle \\
&= \langle a_\parallel, c_\parallel\rangle + \langle a_\parallel, c_\perp\rangle + \langle a_\perp, c_\parallel\rangle + \langle a_\perp, c_\perp\rangle \\
&= |a_\parallel| |c_\parallel| + 0 + 0 + |a_\perp| |c_\perp| \cos \angle(a_\perp,c_\perp) \\
&\ge |a_\parallel| |c_\parallel| - |a_\perp| |c_\perp| \\
&= \frac14 |a| |c| - \frac34 |a| |c| \\
&= -\frac12 |a||c|.
\end{align}
This tells us that $\cos \theta_3 = \frac{\langle a,c\rangle}{|a||c|} \ge -\frac12$. Since $\cos$ is monotone decreasing on $[0,\pi]$, we know that $\theta_3 \le \arccos (-\frac12) = \frac{2\pi}{3}$. (We have equality exactly when $\cos \angle(a_\perp, c_\perp) = -1$: when $a_\perp$ and $c_\perp$ point in parallel but opposite directions.)
By taking $a=c$, we can also see that $\theta_3 = 0$ is achievable, so $0$ is the minimum possible value.
You require the three vectors to satisfy these equations:
\begin{align}
\vec a &= -\lambda_1(\vec b + \vec c), \tag1\\
\vec b &= -\lambda_2(\vec a + \vec c), \tag2\\
\vec c &= -\lambda_3(\vec a + \vec b). \tag3
\end{align}
Use Equations $(2)$ and $(3)$ to substitute for $\vec b$ and $\vec c$
in Equation $(1)$:
$$
\vec a = -\lambda_1(-\lambda_2(\vec a + \vec c) - \lambda_3(\vec a + \vec b)).
$$
Multiply through on the right and collect all the $\vec a$ terms on the left:
$$
(1 - \lambda_1\lambda_2 - \lambda_1\lambda_3)\vec a
= \lambda_1\lambda_2 \vec c + \lambda_1\lambda_3 \vec b.
$$
Divide by the coefficient of $\vec a$:
$$
\vec a
= \frac{\lambda_1\lambda_2}{1 - \lambda_1\lambda_2 - \lambda_1\lambda_3} \vec c
+ \frac{\lambda_1\lambda_3}{1 - \lambda_1\lambda_2 - \lambda_1\lambda_3} \vec b.
\tag4
$$
Since we have not yet established whether $\vec b$ and $\vec c$ are an independent set of vectors, let's consider two cases.
First consider the case where $\vec b$ and $\vec c$ are an independent set of vectors. In that case $\vec a$ can be expressed as a linear combination of $\vec b$ and $\vec c$ in only one way. Therefore the coefficients of $\vec b$ and $\vec c$ in Equation $(1)$ must match the coefficients of $\vec b$ and $\vec c$ in Equation $(4)$:
$$
\frac{\lambda_1\lambda_2}{1 - \lambda_1\lambda_2 - \lambda_1\lambda_3}
= \frac{\lambda_1\lambda_3}{1 - \lambda_1\lambda_2 - \lambda_1\lambda_3}
= -\lambda_1.
$$
This implies that $\lambda_2 = \lambda_3.$
Applying a similar procedure by substituting Equations $(1)$ and $(3)$ into Equation $(2)$, we find that $\lambda_1 = \lambda_3.$
Therefore
$$ \lambda_1 = \lambda_2 = \lambda_3.$$
So we can write $\lambda_1$ in place of $\lambda_2$ and $\lambda_3$, and Equation $(4)$ then says that
$$
\frac{\lambda_1^2}{1 - 2\lambda_1^2} = -\lambda_1.
$$
Using the given assumption that $\lambda_1 > 0$,
$$
\frac{\lambda_1}{1 - 2\lambda_1^2} = -1.
$$
Multiplying by $1 - 2\lambda_1^2$ on both sides and collecting terms,
$$ 2\lambda_1^2 - \lambda_1 - 1 = 0. $$
This equation has two solutions: $\lambda_1 = -\frac12,$ which we must reject because we require that $\lambda_1 > 0,$ and $\lambda_1 = 1.$
Therefore
$$ \lambda_1 = \lambda_2 = \lambda_3 = 1.$$
In short, Equations $(1)$, $(2)$, and $(3)$ are all equivalent to the same equation:
$$ \vec a + \vec b + \vec c = 0. \tag5$$
Now come the bad news. You can set $\vec b$ and $\vec c$ to any two linearly independent vectors in $\mathbb R^2,$ and it follows that
$\vec a = -(\vec b + \vec c)$ provides a non-trivial solution to Equation $(5)$.
On the other hand, suppose $\vec b$ and $\vec c$ are not an independent set of vectors. That is, suppose $\vec b = \mu \vec c.$
Then Equation $(1)$ states that
$$ \vec a = -\lambda_1(\mu + 1) \vec c, $$
that is, the three vectors $\vec a$, $\vec b$, and $\vec c$
all lie in one line.
Equations $(1)$, $(2)$, and $(3)$ are again easily satisfied if
$\lambda_1 = \lambda_2 = \lambda_3 = 1.$
But there are additional solutions, for example,
\begin{align}
\vec a &= \vec b, \\
\vec c &= -3 \vec a = -3 \vec b,\\
\lambda_1 &= \frac12, \\
\lambda_2 &= \frac12, \\
\lambda_3 &= \frac32.
\end{align}
So we cannot really conclude very much about the angles between the vectors;
in the case where the vectors do not all lie in one line,
Equation $(5)$ imposes a few limitations such as any two angles must add up to more than $180$ degrees, but there is nothing like a requirement that any angle be $120$ degrees.
If you make the further requirement that the vectors all have equal magnitude,
then the three vectors actually do have to make $120$-degree angles with each other.
But if you have four equal vectors, then it is sufficient for them to be arranged in two opposite pairs, that is, $\vec a$ and $\vec c$ can be $180$ degrees apart,
and $\vec b$ and $\vec d$ can be $180$ degrees apart,
but $\vec a$ and $\vec b$ can make any angle you want.
Best Answer
If the three vectors lie in the same plane, then obviously any two of them form an angle of $120°$.
If they are not coplanar, such as unit vectors $\vec{OA}$, $\vec{OB}$ and $\vec{OC}$ in diagram below, let $\angle AOB=\angle BOC=\angle COA=\theta$ and let $H$ be the projection of $O$ onto plane $ABC$. By symmetry we have $\angle AHB=\angle BHC=\angle CHA=\theta'$ and moreover $\theta<\theta'$. To justify the last statement notice for instance that $$ AB=2\,OA\sin{\theta\over2}=2\,HA\sin{\theta'\over2}, $$ and $OA>HA$ entails $\theta<\theta'$.
The largest angle is then obtained when all three vectors are coplanar.
This obviously generalises to $n$ dimensions: you can have at most $n+1$ vectors all making the same angles one another. Their coordinates, and thus the angle they form, can be found as described in the answers to this question.