Solution:

The point $M$ ($M \notin \{A, B, C\}$) can lie either on one of the given lines, or in one of the seven regions into which the plane of the triangle $ABC$ is divided by the lines $AB$, $BC$, $CA$.

It is always
$$\begin{aligned} & \frac{MD}{AD} = \frac{MP}{h_a} = \frac{0,5 \cdot MP \cdot BC}{0,5 \cdot h_a \cdot BC} = \frac{F(MBC)}{F(ABC)}, \\ & \text{ (intercept theorem) } \end{aligned}$$
where $P \in BC$ and $MP \perp BC$ and $F(XYZ)$ is the area of triangle $XYZ$. Similar relations hold for the other ratios.

If $M$ lies in the interior or on the boundary of triangle $ABC$, then we have:
$$\frac{MD}{AD} + \frac{ME}{BE} + \frac{MF}{CF} = \frac{F(MBC)}{F(ABC)} + \frac{F(MCA)}{F(ABC)} + \frac{F(MAB)}{F(ABC)} = 1$$
from which $\varepsilon_{1} = \varepsilon_{2} = \varepsilon_{3} = 1$ follows.

Similar considerations also lead to the goal when $M$ lies outside triangle $ABC$, except that $\varepsilon_{1}$, $\varepsilon_{2}$ or $\varepsilon_{3} = -1$, if $M$ lies in region I, II or III respectively. Should $M$ lie in regions IV, V or VI, then exactly two of the $\varepsilon_{i}$ are equal to $-1$.