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(B) Given $a+b+c=21$ and $a, b, c$ are in $AP$,we have $2b = a+c$.Substituting $a+c = 2b$ into the sum equation: $2b + b = 21$ $\Rightarrow 3b = 21$ $

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$7$

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(B) Given $a+b+c=21$ and $a, b, c$ are in $AP$,we have $2b = a+c$.Substituting $a+c = 2b$ into the sum equation: $2b + b = 21$ $\Rightarrow 3b = 21$ $\Rightarrow b = 7$.Since $a, b, c$ are in $AP$,let the common difference be $d$. Then $a = 7-d$,$b = 7$,and $c = 7+d$.Since $a, b, c \in \mathbb{N}$,we must have $a \geq 1$,so $7-d \geq 1 \Rightarrow d \leq 6$.Also,the condition $a \leq b \leq c$ implies $7-d \leq 7 \leq 7+d$,which means $d \geq 0$.Possible values for $d$ are $0, 1, 2, 3, 4, 5, 6$.For each $d$,we get a triplet $(7-d, 7, 7+d)$:If $d=0: (7, 7, 7)$If $d=1: (6, 7, 8)$If $d=2: (5, 7, 9)$If $d=3: (4, 7, 10)$If $d=4: (3, 7, 11)$If $d=5: (2, 7, 12)$If $d=6: (1, 7, 13)$There are $7$ such triplets.

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