Let's list the multiples of each number. Then, we find the smallest number that appears in both lists:

• Multiples of $2$: $\quad 2, 4, 6, {\color{blue}8}, \ldots$

• Multiples of $8$: $\quad {\color{blue}8}, 16, 24, 32, \ldots$

We conclude that the least common multiple of $2$ and $8$ is ${\color{blue}{8}}.$

Let's list the multiples of each number. Then, we find the smallest number that appears in both lists:

• Multiples of $2$: $\quad 2, 4, 6, 8, 10, 12, {\color{blue}14}, \ldots$

• Multiples of $7$: $\quad 7, {\color{blue}14}, 21, \ldots$

We conclude that the least common multiple of $2$ and $7$ is ${\color{blue}{14}}.$

Let's list the multiples of each number. Then, we find the smallest number that appears in both lists:

• Multiples of $12$: $\quad 12, 24, {\color{blue}36}, 48, ...$

• Multiples of $9$: $\quad 9, 18, 27, {\color{blue}36}, 45, ...$

We conclude that the least common multiple of $12$ and $9$ is ${\color{blue}{36}}.$

Let's list the multiples of each number. Then, we find the smallest number that appears in both lists:

• Multiples of $4$: $\quad 4, 8, 12, 16, 20, 24,{\color{blue}28}, 32, \ldots$

• Multiples of $14$: $\quad 14, {\color{blue}28}, 42,\ldots$

We conclude that the least common multiple of $4$ and $14$ is $\bbox[3pt,Gainsboro]{\color{blue}28}.$

Let's list the multiples of each number. Then, we find the smallest number that appears in both lists:

• Multiples of $11$: $\quad 11, 22, {\color{blue}33}, 44,\ldots$

• Multiples of $3$: $\quad 3,6,9,12,15,18,21,24,27,30,{\color{blue}33}, 36, \ldots$

We conclude that the least common multiple of $11$ and $3$ is $\bbox[3pt,Gainsboro]{\color{blue}33}.$

We can solve this problem by finding the least common multiple of $10$ and $6.$

Let's list the multiples of $10$ and $6.$ Then, we find the smallest number that appears in both lists:

• Multiples of $10$: $\quad 10, 20, {\color{blue}30}, 40 ...$

• Multiples of $6$: $\quad 6, 12, 18, 24, {\color{blue}30}, 36, ...$

The least common multiple of $10$ and $6$ is ${\color{blue}{30}}.$ That means that the least number of plates she needs to buy is $\color{blue}30$, because:

• $3$ packages with $10$ forks in each packet gives ${\color{blue}30}$ forks in total.

• $5$ packages $6$ plates in each packet gives ${\color{blue}30}$ plates in total, which is equal to the number of forks.

So the smallest number of plates that Joan needs to buy is $\color{blue}30.$

We can solve this problem by finding the least common multiple of $7$ and $4.$

Let's list the multiples of $7$ and $4.$ Then, we find the smallest number that appears in both lists:

• Multiples of $7$: $\quad 7, 14, 21, {\color{blue}28}, 35, \ldots$

• Multiples of $4$: $\quad 4, 8, 12, 16, 20, 24, {\color{blue}28}, 32, \ldots$

The least common multiple of $7$ and $4$ is ${\color{blue}{28}}.$ Therefore, both John and Carol should pack $28$ bars each.

Let's list the multiples of each number. Then, we find the smallest number that appears in all three lists:

• Multiples of $6$: $\quad 6,12,18,{\color{blue}24},30, \ldots$

• Multiples of $8$: $\quad 8, 16, {\color{blue}24}, 32, \ldots$

• Multiples of $12$: $\quad 12, {\color{blue}24}, 36, \ldots$

We conclude that the least common multiple of $6,$ $8$ and $12$ is ${\color{blue}{24}}.$

Let's list the multiples of each number. Then, we find the smallest number that appears in all three lists:

• Multiples of $7$: $\quad 7, 14, 21, 28, 35, 42, 49, 56, {\color{blue}63}, 70, \ldots$

• Multiples of $9$: $\quad 9, 18, 27, 36, 45, 54, {\color{blue}63}, 72, \ldots$

• Multiples of $21$: $\quad 21, 42, {\color{blue}63}, 84, \ldots$

We conclude that the least common multiple of $7,$ $9,$ and $21$ is $\bbox[3pt, Gainsboro]{\color{blue}63}.$