由题意，可得  $${{(a-b)}^{2}}+{{(a+b)}^{2}}+{{(b-c)}^{2}}+{{(b+c)}^{2}}+{{(a-c)}^{2}}+{{(a+c)}^{2}}$$  $$={{a}^{2}}-2ab+{{b}^{2}}+{{a}^{2}}+2ab+{{b}^{2}}+{{b}^{2}}-2bc+{{c}^{2}}+{{b}^{2}}+2bc+{{c}^{2}}+{{a}^{2}}-2ac+{{c}^{2}}+{{a}^{2}}+2ac+{{c}^{2}}$$  $$=4({{a}^{2}}+{{b}^{2}}+{{c}^{2}})$$，  所以$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}$$  $$=\frac{{{42}^{2}}+{{45}^{2}}+{{64}^{2}}+{{87}^{2}}+{{109}^{2}}+{{151}^{2}}}{4}$$  $$=12534$$． 
有$$3$$个正整数$$a$$，$$b$$，$$c$$，并且$$a\textgreater b\textgreater c$$．从中任取$$2$$个，有$$3$$种不同的取法．将每一种取法取出的$$2$$个数分别作和及作差，得到如下$$6$$个数：$$42$$，$$45$$，$$64$$，$$87$$，$$109$$，$$151$$．则$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}$$的值是（~ ）．

a. $$12532$$ 
b. $$12533$$ 
c. $$12534$$ 
d. $$12535$$ 