Consider a triangle $ABC$ with $BC = 3$. Choose a point $D$ on $BC$ such that $BD = 2$. Find the value of
\[AB^2 + 2AC^2 - 3AD^2.\]
1. **Set up the coordinate system:**
   - Place the origin at point \( E \), the foot of the altitude from \( A \).
   - Let \( A = (0, a) \), \( B = (b, 0) \), and \( C = (3-b, 0) \).
   - Point \( D \) is on \( BC \) such that \( BD = 2 \), so \( D = (2-b, 0) \).

2. **Calculate \( AB^2 \):**
   \[
   AB^2 = (0 - b)^2 + (a - 0)^2 = b^2 + a^2
   \]

3. **Calculate \( 2AC^2 \):**
   \[
   AC^2 = (0 - (3-b))^2 + (a - 0)^2 = (3-b)^2 + a^2
   \]
   \[
   2AC^2 = 2((3-b)^2 + a^2) = 2(a^2 + (3-b)^2) = 2a^2 + 2(9 - 6b + b^2) = 2a^2 + 2b^2 - 12b + 18
   \]

4. **Calculate \( -3AD^2 \):**
   \[
   AD^2 = (0 - (2-b))^2 + (a - 0)^2 = (2-b)^2 + a^2
   \]
   \[
   -3AD^2 = -3((2-b)^2 + a^2) = -3(a^2 + (2-b)^2) = -3a^2 - 3(4 - 4b + b^2) = -3a^2 - 3b^2 + 12b - 12
   \]

5. **Combine the results:**
   \[
   AB^2 + 2AC^2 - 3AD^2 = (a^2 + b^2) + (2a^2 + 2b^2 - 12b + 18) - (3a^2 + 3b^2 - 12b - 12)
   \]
   Simplify the expression:
   \[
   AB^2 + 2AC^2 - 3AD^2 = a^2 + b^2 + 2a^2 + 2b^2 - 12b + 18 - 3a^2 - 3b^2 + 12b - 12
   \]
   \[
   = (a^2 + 2a^2 - 3a^2) + (b^2 + 2b^2 - 3b^2) + (-12b + 12b) + (18 - 12)
   \]
   \[
   = 0 + 0 + 0 + 6
   \]
   \[
   = 6
   \]

The final answer is \(\boxed{6}\).