$b$ is a linear combination of $a_1, a_2, a_3$.
Work Step by Step
$b=2a_1+3a_2$. To solve this, we first construct the top value of $b$, 2, out of our available 1, 0, and 5. We know we want something even and comparatively small, so $2a_1$ is the most simple construction. Next, we try to make the second position of $b$, 11, out of -2, 5, and 0 without disturbing our creation of the 2. Because the top value of $a_2$ is 0, we can use 2(-2)+3(1)=-1, as desired. To check our solution, we look at the third position of $b$, 6, and see that we can construct it by 2(0)+3(2). Thus, $b$ is a linear combination of $a_1, a_2, a_3$ because $b=2a_1+3a_2$.