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Hint: Here, we will first assume the three consecutive multiples are \[3x\], \[3\left( {x + 1} \right)\] and \[3\left( {x + 2} \right)\]. Take these three consecutive multiples, and then use the given conditions to find the required value.

Given that the sum of the three consecutive multiples of 3 is 72.

Let us assume that the three consecutive multiples of 3 are \[3x\], \[3\left( {x + 1} \right)\] and \[3\left( {x + 2} \right)\].

Simplifying the above consecutive multiples, we get

\[3x\]

\[3x + 3\]

\[3x + 6\]

Now as all these three consecutive numbers are multiples of the number three.

Also, we are given that the sum of these three consecutive multiples is 72, so we have

\[3x + 3x + 3 + 3x + 6 = 72\]

Combining the like term in the above equation, we get

\[ \Rightarrow 9x + 9 = 72\]

Subtracting the above equation by 9 on each of the sides, we get

\[

\Rightarrow 9x + 9 - 9 = 72 - 9 \\

\Rightarrow 9x = 63 \\

\]

Dividing the above equation by 9 on each of the sides, we get

\[

\Rightarrow \dfrac{{9x}}{9} = \dfrac{{63}}{9} \\

\Rightarrow x = 7 \\

\]

Substituting this value of \[x\] in the multiple \[3x + 6\] to find the largest multiple of 3, we get

\[

\Rightarrow 3\left( 7 \right) + 6 \\

\Rightarrow 21 + 6 \\

\Rightarrow 27 \\

\]

Thus, the largest multiple of 3 is 27.

Hence, the option C is correct.

Note: In solving these types of questions, you should be familiar with the concept of consecutive numbers. Students should also assume the required consecutive numbers carefully after using the given conditions. If the first term is known then the other terms can be found using the formula of term \[n\]th of an A.P. or directly adding the common difference to the consecutive previous term. Also, we are supposed to write the values properly to avoid any miscalculation.

__Complete step-by-step answer:__Given that the sum of the three consecutive multiples of 3 is 72.

Let us assume that the three consecutive multiples of 3 are \[3x\], \[3\left( {x + 1} \right)\] and \[3\left( {x + 2} \right)\].

Simplifying the above consecutive multiples, we get

\[3x\]

\[3x + 3\]

\[3x + 6\]

Now as all these three consecutive numbers are multiples of the number three.

Also, we are given that the sum of these three consecutive multiples is 72, so we have

\[3x + 3x + 3 + 3x + 6 = 72\]

Combining the like term in the above equation, we get

\[ \Rightarrow 9x + 9 = 72\]

Subtracting the above equation by 9 on each of the sides, we get

\[

\Rightarrow 9x + 9 - 9 = 72 - 9 \\

\Rightarrow 9x = 63 \\

\]

Dividing the above equation by 9 on each of the sides, we get

\[

\Rightarrow \dfrac{{9x}}{9} = \dfrac{{63}}{9} \\

\Rightarrow x = 7 \\

\]

Substituting this value of \[x\] in the multiple \[3x + 6\] to find the largest multiple of 3, we get

\[

\Rightarrow 3\left( 7 \right) + 6 \\

\Rightarrow 21 + 6 \\

\Rightarrow 27 \\

\]

Thus, the largest multiple of 3 is 27.

Hence, the option C is correct.

Note: In solving these types of questions, you should be familiar with the concept of consecutive numbers. Students should also assume the required consecutive numbers carefully after using the given conditions. If the first term is known then the other terms can be found using the formula of term \[n\]th of an A.P. or directly adding the common difference to the consecutive previous term. Also, we are supposed to write the values properly to avoid any miscalculation.

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