The set Φ consists of a finite number of points on the plane. The distance between any two points from Φ is at least $\sqrt{2}$. It is known that a regular triangle with side lenght $3$ cut out of paper can cover all points of Φ. What is the greatest number of points that Φ can consist of?
1. **Determine the area of the equilateral triangle:**
   The side length of the equilateral triangle is given as \(3\). The formula for the area of an equilateral triangle with side length \(a\) is:
   \[
   A = \frac{\sqrt{3}}{4} a^2
   \]
   Substituting \(a = 3\):
   \[
   A = \frac{\sqrt{3}}{4} \times 3^2 = \frac{\sqrt{3}}{4} \times 9 = \frac{9\sqrt{3}}{4}
   \]

2. **Determine the area each point occupies:**
   Since the distance between any two points is at least \(\sqrt{2}\), we can consider each point to occupy a circular area with radius \(\frac{\sqrt{2}}{2}\) (half the minimum distance). The area of such a circle is:
   \[
   A_{\text{point}} = \pi \left(\frac{\sqrt{2}}{2}\right)^2 = \pi \left(\frac{2}{4}\right) = \frac{\pi}{2}
   \]

3. **Calculate the maximum number of points:**
   To find the maximum number of points that can fit within the area of the equilateral triangle, we divide the area of the triangle by the area occupied by each point:
   \[
   \text{Number of points} = \frac{A_{\text{triangle}}}{A_{\text{point}}} = \frac{\frac{9\sqrt{3}}{4}}{\frac{\pi}{2}} = \frac{9\sqrt{3}}{4} \times \frac{2}{\pi} = \frac{9\sqrt{3}}{2\pi}
   \]

4. **Approximate the value:**
   Using the approximate values \(\sqrt{3} \approx 1.732\) and \(\pi \approx 3.1416\):
   \[
   \frac{9 \times 1.732}{2 \times 3.1416} \approx \frac{15.588}{6.2832} \approx 2.48
   \]
   Since the number of points must be an integer, we take the floor of this value:
   \[
   \lfloor 2.48 \rfloor = 2
   \]

5. **Verify the configuration:**
   We need to check if it is possible to place 3 points within the triangle such that the distance between any two points is at least \(\sqrt{2}\). Consider the vertices of the equilateral triangle. The distance between any two vertices of an equilateral triangle with side length 3 is exactly 3, which is greater than \(\sqrt{2}\). Therefore, it is possible to place 3 points in this configuration.

Conclusion:
The greatest number of points that \(\Phi\) can consist of is 3.

The final answer is \(\boxed{3}\)