6. A space probe, moving in a straight line at a constant speed, flies past Mars and measures the distance to this planet every day at exactly 12:00. It is known that on February 1st, the distance was 5 million km, on February 10th - 2 million km, and on February 13th - 3 million km.

Determine when the probe will pass at the minimum distance from Mars.

In this problem, Mars can be considered a point.
Answer: February 9.

Let $d(t)$ be the distance to Mars at time $t$. Suppose Mars is at the origin, the probe starts from the point $\left(x_{0}, y_{0}, z_{0}\right)$, and its velocity vector is $\left(v_{x}, v_{y}, v_{z}\right)$.

Then at time $t$, the probe will be at the point $\left(x_{0}+t \cdot v_{x}, y_{0}+t \cdot v_{y}, z_{0}+t \cdot v_{z}\right)$, which is at a distance of $d(t)=\sqrt{\left(\left(x_{0}+t \cdot v_{x}\right)^{2}+\left(y_{0}+t \cdot v_{y}\right)^{2}+\left(z_{0}+t \cdot v_{z}\right)^{2}\right)}$ from Mars. Squaring this, we note that $D(t)=d^{2}(t)$ is a quadratic function of $t$.

Write $D(t)$ in the form $D(t)=a t^{2}+b t+c$ and find the coefficients $a, b, c$.

To simplify the calculations, introduce a time scale $t$ such that the moment $t=0$ corresponds to February 10, then February 1 corresponds to $t=-9$ and February 13 corresponds to $t=3$.

Substitute the known values of the function $D: D(-9)=25, D(0)=4, D(3)=9$ and solve the resulting system:

$$
\left\{\begin{array} { c } 
{ 8 1 a - 9 b + c = 2 5 } \\
{ c = 4 } \\
{ 9 a + 3 b + c = 9 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a=\frac{1}{3} \\
b=\frac{2}{3} \\
c=4
\end{array}\right.\right.
$$

Find the vertex of the parabola (the graph of the function $\left.D(t)=\frac{1}{3} t^{2}-\frac{2}{3} t+4\right)$:

This is the point $t_{0}=-\frac{b}{2 a}=-1$, which is the point of minimum of the function $D(t)$, corresponding to February 9.