To solve the given problem step by step, we will break it down into two parts as specified in the question.
### Part (a): Find the projection of \( \vec{BC} \) on \( \vec{AB} \)
1. **Determine the coordinates of points A, B, and C**:
- \( A(2, 3, 4) \)
- \( B(4, 3, 2) \)
- \( C(5, 2, -1) \)
2. **Find the vectors \( \vec{AB} \) and \( \vec{BC} \)**:
- \( \vec{AB} = \vec{B} - \vec{A} = (4 - 2, 3 - 3, 2 - 4) = (2, 0, -2) \)
- \( \vec{BC} = \vec{C} - \vec{B} = (5 - 4, 2 - 3, -1 - 2) = (1, -1, -3) \)
3. **Calculate the magnitude of \( \vec{AB} \)**:
\[
|\vec{AB}| = \sqrt{(2)^2 + (0)^2 + (-2)^2} = \sqrt{4 + 0 + 4} = \sqrt{8}
\]
4. **Find the dot product \( \vec{BC} \cdot \vec{AB} \)**:
\[
\vec{BC} \cdot \vec{AB} = (1)(2) + (-1)(0) + (-3)(-2) = 2 + 0 + 6 = 8
\]
5. **Calculate the projection of \( \vec{BC} \) on \( \vec{AB} \)**:
\[
\text{Projection of } \vec{BC} \text{ on } \vec{AB} = \frac{\vec{BC} \cdot \vec{AB}}{|\vec{AB}|} = \frac{8}{\sqrt{8}} = \sqrt{8}
\]
### Part (b): Find the area of triangle ABC
1. **Use the formula for the area of triangle formed by two vectors**:
\[
\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{BC}|
\]
2. **Set up the determinant for the cross product \( \vec{AB} \times \vec{BC} \)**:
\[
\vec{AB} = (2, 0, -2), \quad \vec{BC} = (1, -1, -3)
\]
\[
\vec{AB} \times \vec{BC} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & 0 & -2 \\
1 & -1 & -3
\end{vmatrix}
\]
3. **Calculate the determinant**:
- Expanding the determinant:
\[
= \hat{i} \begin{vmatrix} 0 & -2 \\ -1 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -2 \\ 1 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 0 \\ 1 & -1 \end{vmatrix}
\]
\[
= \hat{i} (0 \cdot -3 - (-2) \cdot -1) - \hat{j} (2 \cdot -3 - (-2) \cdot 1) + \hat{k} (2 \cdot -1 - 0 \cdot 1)
\]
\[
= \hat{i} (0 - 2) - \hat{j} (-6 + 2) + \hat{k} (-2)
\]
\[
= -2\hat{i} + 4\hat{j} - 2\hat{k}
\]
4. **Find the magnitude of the cross product**:
\[
|\vec{AB} \times \vec{BC}| = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}
\]
5. **Calculate the area of triangle ABC**:
\[
\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{BC}| = \frac{1}{2} \cdot 2\sqrt{6} = \sqrt{6}
\]
### Final Answers:
- (a) The projection of \( \vec{BC} \) on \( \vec{AB} \) is \( \sqrt{8} \).
- (b) The area of triangle ABC is \( \sqrt{6} \) square units.
