1. The problem asks to find the value of $a$ so that the parameter curve $\vec{s}(t) = \begin{pmatrix} t^2 - 5t + a \\ t + 1 \end{pmatrix}$ passes through the point $P(0,2)$. 2. To solve this, we use the fact that for some $t$, the vector function equals the point $P(0,2)$, so: $$\begin{cases} t^2 - 5t + a = 0 \\ t + 1 = 2 \end{cases}$$ 3. From the second equation, solve for $t$: $$t + 1 = 2 \implies t = 1$$ 4. Substitute $t=1$ into the first equation: $$1^2 - 5 \cdot 1 + a = 0 \implies 1 - 5 + a = 0$$ 5. Simplify and solve for $a$: $$-4 + a = 0 \implies a = 4$$ 6. Therefore, the value of $a$ that makes the curve pass through $P(0,2)$ is $\boxed{4}$.