We have, $\cos \alpha=\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}, \cos \beta=\frac{\mathbf{b} \cdot \mathbf{c}}{|\mathbf{b}||\mathbf{c}|}$ and $\cos \gamma=\frac{\mathbf{c} \cdot \mathbf{a}}{|\mathbf{c}||\mathbf{a}|}$

Also, it is given that $|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=\lambda$ (say), where $\lambda>0$.

Then, $\cos \alpha=\frac{\mathbf{a} \cdot \mathbf{b}}{\lambda^2} ; \cos \beta=\frac{\mathbf{b} \cdot \mathbf{c}}{\lambda^2}$ and $\cos \gamma=\frac{\mathbf{c} \cdot \mathbf{a}}{\lambda^2}$ $\Rightarrow \cos \alpha+\cos \beta+\cos \gamma=\frac{1}{\lambda^2}$


We know that,
$$
\begin{aligned}
& |\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=(\mathbf{a}+\mathbf{b}+\mathbf{c}) \cdot(\mathbf{a}+\mathbf{b}+\mathbf{c}) \\
= & {\left[|\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2+2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a})\right] } \\
\Rightarrow & {\left[\lambda^2+\lambda^2+\lambda^2+2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a})\right] \geq 0 } \\
& \quad\left[\because|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2 \geq 0\right] \\
\Rightarrow & 2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}) \geq-3 \lambda^2
\end{aligned}
$$

Now, from Eqs. (i) and (ii), we get
$$
\begin{aligned}
\cos \alpha+\cos \beta+\cos \gamma & \geq \frac{1}{\lambda^2}\left(-\frac{3}{2} \lambda^2\right) \\
& =-\frac{3}{2}
\end{aligned}
$$

Thus, the minimum value of $\cos \alpha+\cos \beta+\cos \gamma$ is $-\frac{3}{2}$.