Triangle $ABC$ has a right angle at $C$, and $D$ is the foot of the altitude from $C$ to $AB$. Points $L, M,$ and $N$ are the midpoints of segments $AD, DC,$ and $CA,$ respectively. If $CL = 7$ and $BM = 12,$ compute $BN^2$.
1. **Identify the given information and setup the problem:**
   - Triangle \(ABC\) is a right triangle with \(\angle C = 90^\circ\).
   - \(D\) is the foot of the altitude from \(C\) to \(AB\).
   - Points \(L\), \(M\), and \(N\) are the midpoints of segments \(AD\), \(DC\), and \(CA\), respectively.
   - Given \(CL = 7\) and \(BM = 12\).

2. **Define variables and relationships:**
   - Let \(AL = LD = a\).
   - Since \(L\) is the midpoint of \(AD\), \(AD = 2a\).
   - Since \(M\) is the midpoint of \(DC\), \(MC = \frac{1}{2}DC\).
   - Since \(N\) is the midpoint of \(CA\), \(CN = \frac{1}{2}CA\).

3. **Use the similarity of triangles:**
   - \(\triangle ADC \sim \triangle CBD\) because they share \(\angle C\) and both have right angles.
   - This implies that the ratios of corresponding sides are equal.

4. **Calculate \(AC\) using the Pythagorean theorem in \(\triangle ADC\):**
   - Let \(AC = b\) and \(DC = h\).
   - By the Pythagorean theorem, \(AC^2 + DC^2 = AD^2\).
   - Since \(AD = 2a\), we have \(b^2 + h^2 = (2a)^2\).

5. **Use Stewart's theorem in \(\triangle ADC\):**
   - Stewart's theorem states: \(AC^2 \cdot AD + DC^2 \cdot AD = 2a(a \cdot a + 7^2)\).
   - Substitute \(AC = \frac{2a\sqrt{193}}{7}\) and \(DC = \frac{12}{7}a\):
     \[
     \left(\frac{2a\sqrt{193}}{7}\right)^2 \cdot a + \left(\frac{12}{7}a\right)^2 \cdot a = 2a(a \cdot a + 7^2)
     \]
     \[
     \left(\frac{4a^2 \cdot 193}{49}\right) \cdot a + \left(\frac{144a^2}{49}\right) \cdot a = 2a(a^2 + 49)
     \]
     \[
     \frac{772a^3}{49} + \frac{144a^3}{49} = 2a(a^2 + 49)
     \]
     \[
     \frac{916a^3}{49} = 2a(a^2 + 49)
     \]
     \[
     916a^3 = 98a^3 + 98a
     \]
     \[
     818a^3 = 98a
     \]
     \[
     a^2 = \frac{49}{25}
     \]
     \[
     a = \frac{7}{5}
     \]

6. **Calculate \(DB\):**
   - \(DB = \frac{12}{7} \cdot DC = \frac{12}{7} \cdot \frac{12}{7} \cdot 2 \left(\frac{7}{5}\right) = \frac{288}{25}\).

7. **Calculate \(NM\):**
   - \(NM\) is the midline of \(\triangle ADC\) corresponding to side \(AD = 2a\), so \(NM = a = \frac{7}{5}\).

8. **Calculate \(\cos \angle BNM\):**
   - Since \(NM \parallel AD\), \(\angle BMN = 180^\circ - \angle DBM\).
   - \(\cos \angle BNM = -\cos \angle DBM = -\frac{\frac{288}{25}}{12} = -\frac{24}{25}\).

9. **Apply the Law of Cosines in \(\triangle BMN\):**
   - Using the Law of Cosines:
     \[
     BN^2 = BM^2 + NM^2 - 2 \cdot BM \cdot NM \cdot \cos \angle BNM
     \]
     \[
     BN^2 = 12^2 + \left(\frac{49}{25}\right)^2 - 2 \cdot 12 \cdot \frac{49}{25} \cdot \left(-\frac{24}{25}\right)
     \]
     \[
     BN^2 = 144 + \frac{2401}{625} + \frac{28224}{625}
     \]
     \[
     BN^2 = 144 + \frac{30625}{625}
     \]
     \[
     BN^2 = 144 + 49
     \]
     \[
     BN^2 = 193
     \]

The final answer is \(\boxed{193}\).