Some thoughts.
With the substitution $t = \frac15 + s$, the equation is written as $$20s^3 - 18s^2 + qs - r = 0, \tag{1}$$ where \begin{align*} q &:= \frac{12}{5} - 4 \cos^2 \alpha - 4 \cos^2 \beta - 4 \cos^2 \gamma, \\ r &:= - \frac{9}{25} - \frac15\cos^2 \alpha - \frac15 \cos^2 \beta - \frac15 \cos^2 \gamma + 2 \cos \alpha \cos \beta \cos \gamma. \end{align*}
We need to prove that the cubic equation (1) has at least one real root $s \le 0$.
Currently, we can prove that (1) does not have three positive real roots (see below). (to be continued.)
Assume, for the sake of contradiction, that (1) has three positive real roots. Then we have $q > 0, r > 0$. However, we can prove that $\frac54 r + \frac{3}{16}q \le 0$ that is $$- \cos^2 \alpha - \cos^2 \beta - \cos^2 \gamma + \frac52 \cos \alpha \cos \beta \cos \gamma \le 0. \tag{2}$$ (2) is true since $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma \ge 3\sqrt[3]{\cos^2 \alpha \cos^2 \beta \cos^2 \gamma} \ge 3|\cos \alpha \cos \beta \cos \gamma| \ge \frac52 \cos \alpha \cos \beta \cos \gamma$. Contradiction.