1. Base Case Verification:
- For $|\mathcal{L}| = 1$, there are no triangles formed, hence no discs.
- For $|\mathcal{L}| = 2$, there are still no triangles formed, hence no discs.
- For $|\mathcal{L}| = 3$, exactly one triangle is formed by the three lines, and thus one disc is inscribed in this triangle. The formula $\frac{(|\mathcal{L}|-1)(|\mathcal{L}|-2)}{2}$ gives $\frac{(3-1)(3-2)}{2} = 1$, which matches our observation.

2. Inductive Step:
- Assume the formula holds for $|\mathcal{L}| = n$, i.e., the number of discs intersected by no line in $\mathcal{L}$ is $\frac{(n-1)(n-2)}{2}$.
- Consider $|\mathcal{L}| = n+1$. The $n+1$-th line intersects each of the $n$ lines in $n$ distinct points, creating $n$ segments and two half-lines.
- Each of these $n$ segments creates a new finite region. Thus, the total number of finite regions formed by $n+1$ lines is:
$$n + \frac{(n-1)(n-2)}{2} = \frac{n(n-1)}{2}$$
- Each of these finite regions contains exactly one disc that is inscribed in a triangle formed by three of its sides.

3. Lemma:
- Inside each convex polygon formed by the lines, there is exactly one disc that is tangent to three of its sides.
- Proof of Lemma:
- Start with a small circle tangent to one side $e_1$ of the polygon.
- Homothetically expand the circle until it touches another side $e_2$.
- Let $A$ be the intersection of the lines determined by $e_1$ and $e_2$. Expand the circle from $A$ until it touches a third side $e_3$.
- The circle is now tangent to three sides, forming a triangle $\triangle MNP$.
- If the circle is not the incircle of $\triangle MNP$, it must be an ex-circle. Continue expanding the circle until it touches another side $e_4$.
- This process continues, increasing the radius of the circle until it must stop, at which point the circle is inscribed in a triangle formed by three sides of the polygon.
- To prove uniqueness, assume there are two such discs. The larger disc must be inside the triangle formed by the smaller disc, leading to a contradiction. Hence, there is exactly one such disc.

4. Conclusion:
- The total number of discs intersected by no line in $\mathcal{L}$ is given by the formula $\frac{(|\mathcal{L}|-1)(|\mathcal{L}|-2)}{2}$.

The final answer is $\boxed{\frac{(|\mathcal{L}|-1)(|\mathcal{L}|-2)}{2}}$