**Solution**.

With the substitution $t = \frac15 + s$, the equation is written as
$$f(s) := 20s^3 - 18s^2 + qs - r = 0, \tag{1}$$
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*}

First, let us prove that (1) has three real roots. It suffices to prove that discriminant of $f$ is non-negative, or equivalently
$$f(\cos \alpha, \cos \beta, \cos \gamma) \ge 0,$$
where
\begin{align*}
	f(a, b, c) &:= 64\,{a}^{6}+192\,{a}^{4}{b}^{2}+192\,{a}^{4}{c}^{2}+192\,{a}^{2}{b}^{4
	}\\
&\qquad -156\,{a}^{2}{b}^{2}{c}^{2}+192\,{a}^{2}{c}^{4} +64\,{b}^{6}+192\,{b}^
	{4}{c}^{2}\\
	&\qquad +192\,{b}^{2}{c}^{4} +64\,{c}^{6}-540\,{a}^{3}bc -540\,a{b}^{3
	}c\\
&\qquad -540\,ab{c}^{3}+9\,{a}^{4} +18\,{a}^{2}{b}^{2}+18\,{a}^{2}{c}^{2}+9\,
	{b}^{4}\\
	&\qquad +18\,{b}^{2}{c}^{2}+9\,{c}^{4}+108\,{a}^{2}+108\,{b}^{2}+108\,{
		c}^{2}+27.
\end{align*}
We can prove that $f(a, b, c) \ge 0$ for all real numbers $a, b, c$ (this is verified by *Mathematica*). The desired result follows.

Second, we prove that (1) does not have three positive real roots.
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. Thus, the claim is proved.

Thus, (1) has at least one real root $s \le 0$.

We are done.