1. **State the problem:** Determine if points $A(5,7,4)$ and $B(9,12,4)$ lie on the line $L$ given by the vector equation: $$\vec{r} = \begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix} + \lambda \begin{pmatrix}4 \\ 5 \\ 1\end{pmatrix}$$ 2. **Recall the formula:** A point $P(x,y,z)$ lies on the line if there exists a scalar $\lambda$ such that: $$\begin{cases} x = 1 + 4\lambda \\ y = 2 + 5\lambda \\ z = 3 + 1\lambda \end{cases}$$ 3. **Check point A(5,7,4):** Set up equations: $$5 = 1 + 4\lambda \Rightarrow 4\lambda = 4 \Rightarrow \lambda = 1$$ $$7 = 2 + 5\lambda \Rightarrow 5\lambda = 5 \Rightarrow \lambda = 1$$ $$4 = 3 + 1\lambda \Rightarrow 1\lambda = 1 \Rightarrow \lambda = 1$$ Since all three give $\lambda = 1$, point A lies on the line. 4. **Check point B(9,12,4):** Set up equations: $$9 = 1 + 4\lambda \Rightarrow 4\lambda = 8 \Rightarrow \lambda = 2$$ $$12 = 2 + 5\lambda \Rightarrow 5\lambda = 10 \Rightarrow \lambda = 2$$ $$4 = 3 + 1\lambda \Rightarrow 1\lambda = 1 \Rightarrow \lambda = 1$$ Here, $\lambda$ is $2$ from the first two equations but $1$ from the third, which is a contradiction. Therefore, point B does not lie on the line. **Final answer:** Point $A$ lies on the line $L$, but point $B$ does not.