If \(\vec{A}=a x \hat{\imath}+b y \hat{\jmath}+c z \hat{k}\) where a, b, c are constants, then \(\iint_S \vec{A} . d \vec{S}\) where S is the surface of a unit sphere, is

1
\(\frac{4}{3} \pi(a+b+c)^2\)
2
\(\frac{4}{3} \pi(a+b+c)\)
3
0
4
\(\frac{4}{3} \pi\left(a^2+b^2+c^2\right)\)

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