The minimum value of $a^2+b^2+c^2+\frac1{a^2}+\frac1{b^2}+\frac1{c^2}?$

$a^2+b^2+c^2+a^{-2}+b^{-2}+c^{-2}=(a-a^{-1})^2+(b-b^{-1})^2+(c-c^{-1})^2+6$, whence the minimum occurs when $a=a^{-1},b=b^{-1},c=c^{-1}$ and is $6$.

Might as well take advantage of the fact that it's a multiple choice question.

First, is it possible that the quantity is ever zero? Next, can you find $a, b, c$ such that $$a^2+b^2+c^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} = 6?$$

By the symmetry of the expression, the minimum is attained with $a=b=c$ so we look for the minimum of $$f(x)=3(x^2+x^{-2})$$ By the derivative of the function $f$ we find that the minimum is at $x=1$, hence your answer must be $6$.