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(A) The sum of the ages of $5$ members $3 \, 	ext{years}$ ago was $17 	imes 5 = 85 \, 	ext{years}$.The current sum of the ages of these $5$ members

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(A) The sum of the ages of $5$ members $3 \, 	ext{years}$ ago was $17 	imes 5 = 85 \, 	ext{years}$.The current sum of the ages of these $5$ members is $85 + (3 	imes 5) = 85 + 15 = 100 \, 	ext{years}$.The current average age of $6$ members (including the baby) is $17 \, 	ext{years}$.Therefore,the current total sum of the ages of $6$ members is $17 	imes 6 = 102 \, 	ext{years}$.The age of the new baby is the difference between the total sum of $6$ members and the total sum of the original $5$ members: $102 - 100 = 2 \, 	ext{years}$.

$3 \, \text{years}$ ago,the average age of a family of $5 \, \text{members}$ was $17 \, \text{years}$. With the birth of a new baby,the average age of the $6 \, \text{members}$ remains the same even today. Find the age of the new baby in years.

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