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Hint: Here to proceed the solution we need to know the multiples of 3 and 6. Make set A and set B as per given condition.

Here we are given with two sets where with the condition, Where

A = $\{ x \in {\rm N}:{\text{x is a multiple of }}3\} $

Here set A is a multiple of 3 which are natural numbers.

Then $A = \{ 3,6,9,12,15,18,.......\} $

$B = \{ x \in {\rm N}:{\text{x is a multiple of 6}}\} $

Here set B is a multiple of 6 which are natural numbers.

Then $B = \{ 6,12,18,24,.....\} $

Now we got both the values of set A and set B

Then

$

A - B = \{ 3,6,9,12,15,18,.....\} - \{ 6,12,18,24,30,....\} \\

A - B = \{ 3,9,12,15,21,.....\} \\

$

Here A-B means we have removed the element of set B from set A.

Option C is the correct answer.

NOTE: In these problems first we have to find the value of set A by given condition and in the same way we have to find the values of set B .Later we have to find A-B which means we have to subtract values of set B from set A. In this type of problem mainly we have to focus on the conditions given to the sets because different sets have different conditions.

Here we are given with two sets where with the condition, Where

A = $\{ x \in {\rm N}:{\text{x is a multiple of }}3\} $

Here set A is a multiple of 3 which are natural numbers.

Then $A = \{ 3,6,9,12,15,18,.......\} $

$B = \{ x \in {\rm N}:{\text{x is a multiple of 6}}\} $

Here set B is a multiple of 6 which are natural numbers.

Then $B = \{ 6,12,18,24,.....\} $

Now we got both the values of set A and set B

Then

$

A - B = \{ 3,6,9,12,15,18,.....\} - \{ 6,12,18,24,30,....\} \\

A - B = \{ 3,9,12,15,21,.....\} \\

$

Here A-B means we have removed the element of set B from set A.

Option C is the correct answer.

NOTE: In these problems first we have to find the value of set A by given condition and in the same way we have to find the values of set B .Later we have to find A-B which means we have to subtract values of set B from set A. In this type of problem mainly we have to focus on the conditions given to the sets because different sets have different conditions.

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