Determine the minimum number of lines that can be drawn on the plane so that they intersect in exactly $200$ distinct points.

(Note that for $3$ distinct points, the minimum number of lines is $3$ and for $4$ distinct points, the minimum is $4$)
1. **Understanding the Problem:**
   We need to find the minimum number of lines that can be drawn on a plane such that they intersect in exactly 200 distinct points.

2. **General Formula for Intersections:**
   For \( n \) lines, the maximum number of intersection points is given by the combination formula:
   \[
   \binom{n}{2} = \frac{n(n-1)}{2}
   \]
   This formula counts the number of ways to choose 2 lines out of \( n \) lines, each pair of which intersects at a unique point.

3. **Finding the Minimum \( n \):**
   We need to find the smallest \( n \) such that:
   \[
   \frac{n(n-1)}{2} \geq 200
   \]
   Solving for \( n \):
   \[
   n(n-1) \geq 400
   \]
   \[
   n^2 - n - 400 \geq 0
   \]
   Solving the quadratic equation \( n^2 - n - 400 = 0 \) using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   n = \frac{1 \pm \sqrt{1 + 1600}}{2} = \frac{1 \pm \sqrt{1601}}{2}
   \]
   Approximating \( \sqrt{1601} \approx 40.01 \):
   \[
   n = \frac{1 + 40.01}{2} \approx 20.505
   \]
   Since \( n \) must be an integer, we round up to the next whole number:
   \[
   n = 21
   \]

4. **Verification:**
   Check if \( n = 21 \) satisfies the condition:
   \[
   \frac{21 \times 20}{2} = 210
   \]
   Since 210 is greater than 200, \( n = 21 \) is a valid solution.

5. **Configuration:**
   The solution provided suggests using 10 pairs of parallel lines and 1 additional line that intersects all other lines. However, this configuration does not match the requirement of exactly 200 distinct intersection points. Instead, we should use the general formula directly.

6. **Conclusion:**
   The minimum number of lines required to achieve exactly 200 distinct intersection points is 21.

The final answer is \( \boxed{ 21 } \)