We are given the vector magnitude $|\vec{a}| = 5$ and the scalar range $-2 \le \lambda \le 1$. We need to find the range of the magnitude of the vector $\lambda\vec{a}$, which is given by $|\lambda\vec{a}| = |\lambda| |\vec{a}|$.
Since $-2 \le \lambda \le 1$, the absolute value $|\lambda|$ represents the distance of $\lambda$ from zero on the number line. The values of $\lambda$ range from $-2$ to $1$. The maximum distance from zero is at $\lambda = -2$, where $|\lambda| = 2$. The minimum distance from zero is at $\lambda = 0$, where $|\lambda| = 0$. Thus, $0 \le |\lambda| \le 2$.
Multiplying the inequality by $|\vec{a}| = 5$: $$0 \times 5 \le |\lambda| |\vec{a}| \le 2 \times 5$$ $$0 \le |\lambda\vec{a}| \le 10$$
The greatest value is $10$ and the smallest value is $0$. The sum is: $$10 + 0 = 10$$
Final Answer: 10
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