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If $a, b, c$ and $d$ are integers such that $a+b+c+d=30$ , then the minimum possible value of $( a-b )^{2}+( a-c )^{2}+( a-d)^{2}$ is

1. $1$
2. $2$
3. $5$
4. $6$

Given that, $a,b,c$ and $d \in \mathbb{Z}$

And, $a+b+c+d =30 \quad \longrightarrow (1)$

We know that, ${(a-b)}^{2}+{(a-c)}^{2}+{(a-d)}^{2} \geq 0$

Lets take, ${(a-b)}^{2}+{(a-c)}^{2}+{(a-d)}^{2} = 0$

$\Rightarrow \boxed{a=b=c=d}$

From equation $(1)$, we get.

$a+a+a+a = 4a$

$\Rightarrow 4a=30$

$\Rightarrow a=\frac{30}{4} = \frac{15}{2} = 7.5$ (not an integer)

We need to distribute $a,b,c$ and $d$ in such a way that, $a+b+c+d =30,$ and $a,b,c$ and $d$ are as closed to $7.5$ as possible

One possible way, $a=7,b=7,c=8,d=8$

So,  ${(a-b)}^{2}+{(a-c)}^{2}+{(a-d)}^{2}= {(7-7)}^{2}+{(7-8)}^{2}+{(7-8)}^{2}$

$\qquad = 0^{2}+(-1)^{2}+(-1)^{2}$

$\qquad = 0+1+1 =2$

$\textbf{PS:}$ We can also take, $a=8,b=7,c=8,d=7.$

$\therefore$ The minimum positive value of ${(a-b)}^{2}+{(a-c)}^{2}+{(a-d)}^{2}$ is $2.$

Correct Answer $:\text{B}$
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