the answer is   (abc)^2/3

we use this property that if ther is 1^infinity form inquestion like lim x (tends to)a   [f(x)]^g(x)

then we can write it as  

e^lim x (tends to)a    [f(x)-1]*g(x)

now  after aplying this property

e^lim x(tends to) 0  ( [a^x+b^x+c^x]/3 -1)2/x

e^lim x(tends to) 0  ( a^x+b^x+c^x-3/3)2/x

e^lim x(tends to) 0  ( a^x-1+b^x-1+c^x-1/3)2/x

 e^lim x(tends to) 0  2/3 [( a^x-1/3)+(b^x-1/3)+(c^x-1/3)]

now apply the property  lim x(tends to 0) a^x-1/x  =loga to base e

 e^lim x(tends to) 0  2/3[log a +log b +log c]

 e^  1/3[logabc]

therefore   ans...(2/3abc)

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